openstax chemistry 2e

OpenStax Chemistry 2e 15.1: Precipitation and Dissolution

of 27

Chemistry 2e 15: Equilibria of Other Reaction Classes

15.1: Precipitation and Dissolution 1. Complete the changes in concentrations for each of the following reactions: (a) AgI( ) Ag ( ) I ( )

_____s aq aq

x

(b) 2 23 3CaCO ( ) Ca ( ) CO ( )

_____ s aq aq

x

(c) 22Mg(OH) ( ) Mg ( ) 2OH ( )

_____s aq aq

x

(d) 2 33 4 2 4Mg (PO ) ( ) 3Mg ( ) 2PO ( )

_____s aq aq

x

(e) 2 3 –5 4 3 4Ca (PO ) OH( ) 5Ca ( ) 3PO ( ) OH ( )

_____ _____ s aq aq aq

x

Solution In dissolution, one unit of substance produces a quantity of discrete ions or polyatomic ions that equals the number of times that the subunit appears in the formula. (a) + –AgI( ) Ag ( ) + I ( )

s aq aq

x x

� ��

Dissolving AgI(s) must produce the same amount of I– ion as it does Ag+ ion. (b) 2+ 2–

3 3CaCO ( ) Ca ( ) + CO ( )

s aq aq

x x

� ��

Dissolving CaCO3(s) must produce the same amount of Ca2+ ion as it does 23CO ion.

(c) 2 –2Mg OH ( ) Mg ( ) 2OH ( )

2

s aq aq

x x

� ��

When one unit of Mg(OH)2 dissolves, two ions of OH– are formed for each Mg2+ ion. (d) 2 3–

3 4 42Mg PO ( ) 3Mg ( ) 2PO ( )

3 2

s aq aq

x x

� ��

One unit of Mg3(PO4)2 provides two units of 34PO ion and three units of Mg2+ ion.

(e) 2 3– –5 4 43Ca PO OH( ) 5Ca ( ) 3PO ( ) OH ( )

5 3

s aq aq aq

x x x

� ��

One unit of Ca5(PO4)3OH dissolves into five units of Ca2+ ion, three units of 34PO ion, and one

unit of OH– ion. 2. Complete the changes in concentrations for each of the following reactions: (a) 2 2

4 4BaSO ( ) Ba ( ) SO ( ) _____

s aq aqx


of 27

(b) 22 4 4Ag SO ( ) 2Ag ( ) SO ( )

_____ s aq aq

x

(c) 33Al(OH) ( ) Al ( ) 3OH ( )

_____s aq aq

x

(d) 2Pb(OH)Cl( ) Pb ( ) OH ( ) Cl ( ) _____ _____

s aq aq aqx

(e) 2 33 4 2 4Ca (AsO ) ( ) 3Ca ( ) 2AsO ( )

3 _____s aq aq

x

Solution In dissolution, one unit of substance produces a quantity of discrete ions or polyatomic ions that equals the number of times that the subunit appears in the formula. (a) 2+ 2

4 4BaSO ( ) Ba ( ) + SO ( )

s aq aq

x x

� ��

(b) + 22 4 4Ag SO ( ) 2Ag ( ) + SO ( )

2

s aq aq

x x

� ��

(c) 3+3Al(OH) ( ) Al ( ) + 3OH ( )

3

s aq aq

x x

� ��

(d) 2+Pb(OH)Cl( ) Pb ( ) + OH ( ) + Cl ( )

s aq aq aq

x x x

� ��

(e) 2+ 33 4 2 4Ca (AsO ) ( ) 3Ca ( ) + 2AsO ( )

3 2

s aq aq

x x

� �� ]

3. How do the concentrations of Ag+ and 24CrO in a saturated solution above 1.0 g of solid

Ag2CrO4 change when 100 g of solid Ag2CrO4 is added to the system? Explain. Solution There is no change. A solid has an activity of 1 whether there is a little or a lot. 4. How do the concentrations of Pb2+ and S2– change when K2S is added to a saturated solution of PbS? Solution The added K2S causes the concentration of S2– to increase and the concentration of Pb2+ to decrease via formation of more PbS by the common ion effect. 5. What additional information do we need to answer the following question: How is the equilibrium of solid silver bromide with a saturated solution of its ions affected when the temperature is raised? Solution The solubility of silver bromide at the new temperature must be known. Normally the solubility increases and some of the solid silver bromide will dissolve. 6. Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: CoSO3, CuI, PbCO3, PbCl2, Tl2S, KClO4?


of 27

Solution CoSO3, PbCO3, Tl2S 7. Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: AgCl, BaSO4, CaF2, Hg2I2, MnCO3, ZnS, PbS? Solution CaF2, MnCO3, and ZnS; each is a salt of a weak acid and the hydronium ion from water reacts with the anion, causing more solid to dissolve to maintain the equilibrium concentration of the anion 8. Write the ionic equation for dissolution and the solubility product (Ksp) expression for each of the following slightly soluble ionic compounds: (a) PbCl2 (b) Ag2S (c) Sr3(PO4)2 (d) SrSO4 Solution (a) 2 – 2 – 2

2 spPbCl ( ) Pb ( ) 2Cl ( ), [Pb ][Cl ]s aq aq K � �� ;

(b) 2– 2 2–2 spAg S( ) 2Ag ( ) S ( ), [Ag ] [S ]s aq aq K � �� ;

(c) 2 3 2 3 3 23 4 2 4 sp 4Sr (PO ) ( ) 3Sr ( ) 2PO ( ), [Sr ] [PO ]s aq aq K � �� ;

(d) 2 2 2 24 4 sp 4SrSO ( ) Sr ( ) SO ( ), [Sr ][SO ]s aq aq K � ��

9. Write the ionic equation for the dissolution and the Ksp expression for each of the following slightly soluble ionic compounds: (a) LaF3 (b) CaCO3 (c) Ag2SO4 (d) Pb(OH)2 Solution (a) 3 – 3 – 3

3 spLaF ( ) La ( ) 3F ( ) [La ][F ]s aq aq K � �� ;

(b) 2 2 2 23 3 sp 3CaCO ( ) Ca ( ) CO ( ) [Ca ][CO ]s aq aq K � �� ;

(c) 2 2 22 4 4 sp 4Ag SO ( ) 2Ag ( ) SO ( ) [Ag ] [SO ]s aq aq K � �� ;

(d) 2 – 2 – 22 spPb(OH) ( ) Pb ( ) 2OH ( ) [Pb ][OH ]s aq aq K � ��

10. The Handbook of Chemistry and Physics (http://hbcponline.com/faces/contents/ContentsSearch.xhtml) gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each. (a) BaSiF6, 0.026 g/100 mL (contains 2

6SiF ions) (b) Ce(IO3)4, 1.5 10–2 g/100 mL (c) Gd2(SO4)3, 3.98 g/100 mL (d) (NH4)2PtBr6, 0.59 g/100 mL (contains 2

6PtBr ions) Solution

http://hbcponline.com/faces/contents/ContentsSearch.xhtml)


of 27

Convert each concentration into molar units. Multiply each concentration by 10 to determine the mass in 1 L, and then divide by the molar mass:

(a) 1

46 1

0.26 g LBaSiF : 9.306 10 279.403 g mol

M

,

K = 2 26[Ba ][SiF ] (9.306 10–4)(9.306 10–4) = 8.7 10–7;

(b) 1

43 4 1

0.15 g LCe(IO ) : 1.79 10 839.726 g mol

M

,

K = 4 43[Ce ][IO ] = (1.79 10–4)(4 1.79 10–4)4 = 4.7 10–17;

(c) 1

22 4 3 1

39.8 g LGd (SO ) : 6.60 10 602.69 g mol

M

,

K = 3 2 2 34[Ga ] [SO ] = (2 6.60 10–2)2(3 6.60 10–2)3 = (0.0174)(0.00776) = 1.35 10–4;

(d) 1

34 6 1

5.9 g L(NH) PtBr : 8.3 10 710.581 g mol

M

,

K = 2 24 6[NH ] [PtBr ] = (2 8.3 10–3)2(8.30 10–3) = (2.76 10–4)(8.3 10–3) = 2.3 10–6

11. The Handbook of Chemistry and Physics (http://hbcponline.com/faces/contents/ContentsSearch.xhtml) gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each. (a) BaSeO4, 0.0118 g/100 mL (b) Ba(BrO3)2H2O, 0.30 g/100 mL (c) NH4MgAsO46H2O, 0.038 g/100 mL (d) La2(MoO4)3, 0.00179 g/100 mL Solution Convert each concentration into molar units. Multiply each concentration by 10 to determine the mass in 1 L, and then divide the molar mass.

(a) BaSeO4: –1

4–1

0.118 g L = 4.21 10 280.28 g mol

M ,

K = [Ba2+] 24[SeO ] = (4.21 10–4)(4.21 10–4) = 1.77 10–7;

(b) Ba(BrO3)2H2O: –1

3–1

3.0 g L = 7.3 10 411.147 g mol

M ,

K = [Ba2+] 23[BrO ] = (7.3 10–3)(2 7.3 10–3)2 = 1.6 10–6;

(c) NH4MgAsO46H2O: –1

3–1

0.38 g L = 1.3 10 289.3544 g mol

M ,

K = 4[NH ] [Mg2+] 34[AsO ] = (1.3 10–3)3 = 2.2 10–9;

(d) La2(MoO4)3: –1

5–1

0.0179 g L = 2.36 10 757.62 g mol

M ,

K = [La3+]2 2 34[MoO ] = (2 2.36 10–5)2(3 2.36 10–5)3 = 2.228 10–9 3.549 10–13 =

7.91 10–22

http://hbcponline.com/faces/contents/ContentsSearch.xhtml)


of 27

12. Use solubility products and predict which of the following salts is the most soluble, in terms of moles per liter, in pure water: CaF2, Hg2Cl2, PbI2, or Sn(OH)2. Solution Since each of the compounds dissociates into three ions upon dissolution, their solubilities may be compared by examining the value of the compounds’ respective solubility products. The larger the value of Ksp, the greater will be the solubility. The Ksp values are as follows: CaF2: Ksp = 4.0 10–11; Hg2Cl2: Ksp = 1.1 10–18; PbI2: Ksp = 1.4 10–8; Sn(OH)2: Ksp = 3 10–27. PbI2 has the largest Ksp and is, therefore, the most soluble in terms of moles per liter. 13. Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product: (a) KHC4H4O6 (b) PbI2 (c) Ag4[Fe(CN)6], a salt containing the 4

6Fe(CN) ion (d) Hg2I2 Solution Let x be the molar solubility. (a) Ksp = 4 4 6[K ][HC H O ] = 3 10–4 = x2, x = 2 10–2M; (b) Ksp = [Pb2+][I–]2 = 1.4 10–8 = x(2x)3 = 4x3, x = 1.5 10–3M; (c) Ksp = 4 4

6[Ag ] [Fe(CN) ] = 1.55 10–41 = (4x)4x = 256x5, x = 2.27 10–9M; (d) Ksp = 2 2

2[Hg ][I ] = 4.5 10–29 = [x][2x]2 = 4x3, x = 2.2 10–10M 14. Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product: (a) Ag2SO4 (b) PbBr2 (c) AgI (d) CaC2O4H2O Solution Let x be the molar solubility. (a) Ksp = [Ag+]2 2

4[SO ] = 1.2 10–5 = (2x)2(x) = 4x3, x = 1.44 10–2M; (b) Ksp = [Pb2+][Br–]2 = 4.6 10–6 = x(2x)2 = 4x3, x = 1.1 10–2M; (c) Ksp = [Ag+][I–] = 1.5 10–16 = x2, x = 1.2 10–8M; (d) Ksp = [Ca2+] 4 2+

2[C O ] [H2O] = 1.96 10–9 = x2, x = 4.4 10–5M 15. Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected. (a) AgCl(s) in 0.025 M NaCl (b) CaF2(s) in 0.00133 M KF (c) Ag2SO4(s) in 0.500 L of a solution containing 19.50 g of K2SO4 (d) Zn(OH)2(s) in a solution buffered at a pH of 11.45 Solution (a) Ksp = 1.6 10–10 = [Ag+][Cl–] = x(x + 0.025), where x = [Ag+]. Assume that x is small when compared with 0.025 and therefore ignore it:


of 27

1091.6 10 6.4 10 [Ag ], [Cl ] 0.025

0.025x M M

Check: 9

56.4 10 100% = 2.6 10 %0.025

MM

, an insignificant change;

(b) Ksp = 4.0 10–11 = [Ca2+][F–]2 = x(2x + 0.00133 M)2, where x = [Ca2+]. Assume that x is small when compared with 0.0013 M and disregard it:

115 2

24.0 10 2.2 10 [Ca ], [F ] 0.0013 (0.00133)

x M M

Check: 52.26 10 100% = 1.70%

0.00133 M

M

. This value is less than 5% and can be ignored.

(c) Find the concentration of K2SO4:

119.50 g = 0.1119 mol

174.260 g mol

24

0.1119 mol = 0.2238 = [SO ]0.5 L

M

Ksp = 1.12 10–5 = [Ag+]2 24[SO ] = 4x2(x + 0.2238)

52 51.2 10 = = 1.34 10

4(0.2238)x

x = 3.7 10–3[Ag+] = 2x = 7.4 10–3M

Check: 3

23.7 10 100% = 1.64 100.2238

; the condition is satisfied.

(d) Find the concentration of OH– from the pH: pOH = 14.00 – 11.45 = 2.55 [OH–] = 2.8 10–3M Ksp = 4.5 10–17 = [Zn2+][OH–]2 = x(2x + 2.8 10–3)2 Assume that x is small when compared with 2.8 10–3:

1712 2+

3 24.5 10 = = 5.7 10 = [Zn ]

(2.8 10 )x M

Check: 12

73

5.7 10 100% = 2.0 10 %2.8 10

; x is less than 5% of [OH–] and is, therefore,

negligible. In each case the change in initial concentration of the common ion is less than 5%. 16. Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected. (a) TlCl(s) in 1.250 M HCl (b) PbI2(s) in 0.0355 M CaI2 (c) Ag2CrO4(s) in 0.225 L of a solution containing 0.856 g of K2CrO4 (d) Cd(OH)2(s) in a solution buffered at a pH of 10.995 Solution (a) Let x = [Tl+]:


of 27

Ksp = 1.7 10–4 = [Tl+][Cl–] = x(x + 1.250) Assume that x is small in comparison to 1.250 and eliminate it from the expression x + 1.250:

–4

–4 –1.7 10 1.5 10 Tl ; Cl 1.250 1.250

x M M

Check: –41.5 10 0.012%

1.250 100%


(b) Let x = [Pb+]; [I–] = 2 0.0355 = 0.0710 M Ksp = 1.4 10–8 = [Pb+][I–] = x(2x + 0.0710)2 Assume that 2x is small when compared with 0.0710 M.

–8–6

21.4 10 = 2.78 10 Pb

0.0710x M ; [I–] = 0.0710 M

Check: –6

–32.78 10 3.92 10 % 0.0710 100%


(c) Determine the concentration 24[CrO ] . Molar mass K2CrO4 = 155.092 g/mol. In 0.255 L:

–2–1

0.856 g 2.45 10 155.092 g mol

M M .

Let x be the change in concentration of 24CrO :

Ksp = 9 10–12 = [2x]2[x + 2.45 10–2] Assume that x is smaller when compared with 2.45 10–2M: x2 = 9.18 10–11; x = 1 10–5 [Ag+] = 2x = 2 10–5M

24[CrO ] = 1 10–5 + 2.45 10–2 = 2.45 10–2M

The assumption is obviously valid. (d) Let x = [Cd2+]. Calculate the buffered [OH–] concentration from the pH: pH = 10.995, pOH = 14.000 – pH = 3.005 [OH–] = 9.8855 10–4M Ksp = 5.9 10–15 = [Cd2+][OH–]2 = x(2x + [OH–])2 Ksp = 5.9 10–15 = [Cd2+][OH–]2 = x(2x + 9.886 10–4)2 Assume that x is small when compared with 9.886 10–4M:

159 2+

4 25.9 10 = = 6.0 10 = [Cd ]

(9.886 10 )x M

[OH–] = 9.886 10–4M

Check: –9

–4

6.0 10 100% 0.00061%9.886 10 M


17. Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that it is not appropriate to neglect the changes in the initial concentrations of the common ions. (a) TlCl(s) in 0.025 M TlNO3 (b) BaF2(s) in 0.0313 M KF


of 27

(c) MgC2O4 in 2.250 L of a solution containing 8.156 g of Mg(NO3)2 (d) Ca(OH)2(s) in an unbuffered solution initially with a pH of 12.700 Solution (a) Ksp = 1.7 10–4 = [Ti+][Cl–]; Let x = [Cl–]: 1.7 10–4 = (x = 0.025)x Assume that x is small when compared with 0.025:

431.7 10 = = 6.8 10

0.025x M

Check: 36.8 10 100% = 27%

0.025

This value is too large to drop x. Therefore solve by using the quadratic equation: 2 4

2b b ac

a

x2 + 0.025x – 1.7 10–4 = 0 4 4 30.025 6.25 10 6.8 10 0.025 1.305 10

2 20.025 0.0361 0.0056

2

x

M

(Use only the positive answer for physical sense.) [Ti+] = 0.025 + 0.0056 = 3.1 10–2M [Cl–] = 6.1 10–3 (b) Ksp = 2.4 10–5 = [Ba2+][F–]2; Let x = [Ba2+] If we drop x from x + 0.0313, x would be equal to 7.7 10–4 M

Check: 47.7 10 100% = 2.4%

0.0313

This value is less than 5%, so [Ba2+] = 7.7 10–4M [F–] = 0.0321 M; (c) Find the molar concentration of the Mg(NO3)2. The molar mass of Mg(NO3)2 is 148.3149

g/mol. The number of moles is 18.156 g = 0.05499 mol

148.3149 g mol

0.05499 mol = = 0.02444 2.250 L

M M

Let x = 22 4[C O ] and assume that x is small when compared with 0.02444 M.

Ksp = 7 10–7 = [Mg2+] 22 4[C O ] = (x)(x + 0.02444)

0.02444x = 7 10–7 x = 2

2 4[C O ] = 2.9 10–5

Check: 52.9 10 100% = 0.12%

0.02444

This value is less than 5%, so [Ca2+] = 2.9 10–5 M


of 27

[OH–] = 0.0501 M [Mg2+] = 0.0244 M (d) pH = 12.700; pOH = 1.300 [OH–] = 0.0501 M; Let x = [Ca2+] Ksp = 7.9 10–6 = [Ca2+][OH–]2 = (x)(x + 0.050)2 Assume that x is small when compared with 0.050 M: x = [Ca2+] = 3.15 10–3 (one additional significant figure is carried)

Check: 33.15 10 100% = 6.28%

0.050

This value is greater than 5%, so a more exact method, such as successive approximations, must be used. Begin by choosing the value of x that has just been calculated: x′(3.15 10–3 + 0.0501)2 = 7.9 10–6 or

63

3

7.9 10 = 2.8 102.836 10

x

[Ca2+] = 2.8 10–3M [OH–] = (2.8 10–3 + 0.0501) = 0.053 10–2M In each case, the initial concentration of the common ion changes by more than 5%. 18. Explain why the changes in concentrations of the common ions in Exercise 17 can be neglected. Solution The changes in concentration are less than 5%. 19. Explain why the changes in concentrations of the common ions in Exercise 18 cannot be neglected. Solution The changes in concentration are greater than 5% and thus exceed the maximum value for disregarding the change. 20. Calculate the solubility of aluminum hydroxide, Al(OH)3, in a solution buffered at pH 11.00. Solution At pH 11.00, pOH = 3.00. Then [OH–] = 1.0 10–3M. Ksp = [Al3+][OH–]3 = [Al3+](1 10–3)3 = 1.9 10–33

323+ 23

3 32 10solubility = [Al ] = = 2 10

(1.0 10 )M

21. Refer to Appendix J for solubility products for calcium salts. Determine which of the calcium salts listed is most soluble in moles per liter and which is most soluble in grams per liter. Solution Ca(OH)2: [Ca2+][OH–]2 = 1.3 10–6 Let x be [Ca2+] = molar solubility; then [OH–] = 2x Ksp = x(2x)2 = 4x3 = 1.3 10–6 x3 = 0.069 10–6 x = 0.069 M CaCO3: [Ca2+] 2

3[CO ] = 8.7 10–9 Let x be [Ca2+]; then 2

3[CO ] = [Ca2+] = molar solubility Ksp = x2 = 8.7 10–9 x = 9.3 10–5M


of 27

CaSO4·2H2O: [Ca2+] 24[SO ] �[H2O]2 = 6.1 10–5

Let x be [Ca2+] = molar solubility = 24[SO ] ; then [H2O] = 2x

Ksp = (x)(x)(2x)2 = 6.1 10–5 x4 = 1.53 10–5 x = 0.062 M This value is more than four times the value given by Handbook of Chemistry and Physics (http://www.hbcpnetbase.com/) of (0.014 M) and reflects the complex interaction of water within the precipitate: CaC2O4·H2O: [Ca2+] 2

2 4[C O ] [H2O] = 1.96 × 10–8 Let x be [Ca2+] = molar solubility = 2

2 4[C O ] = [H2O] x3 = 1.96 10–8 x = 2.7 10–3M In this case, the interaction of water is also complex and the solubility is considerably less than that calculated. Ca3(PO4)2: [Ca2+]3 3 2

4[PO ] = 1.3 10–32 Upon solution there are three Ca2+ and two 3

4PO ions. Let the concentration of Ca2+ formed

upon solution be x. Then 23

x is the concentration of 34PO :

23 3 32 52 = 1.3 10 = 0.4444

3x x x x

x = 4.9 10–7M = [Ca2+] The solubility is then one-third the concentration of Ca2+, or 1.6 10–7. CaSO4 2H20 is the most soluble Ca salt. 22. Most barium compounds are very poisonous; however, barium sulfate is often administered internally as an aid in the X-ray examination of the lower intestinal tract (Figure 15.4). This use of BaSO4 is possible because of its low solubility. Calculate the molar solubility of BaSO4 and the mass of barium present in 1.00 L of water saturated with BaSO4. Solution The reaction is:

2 24 4BaSO ( ) Ba ( ) SO ( )s aq aq� ��

There is a 1:1 ratio of ions generated, so that [Ba2+] = 24[SO ] in solution. The solubility product

is: Ksp = [Ba2+] 2

4[SO ] = 2.3 10–8 Let x be the change in concentration of Ba2+. Before dissolving some BaSO4, the concentration of Ba2+ is 0. At equilibrium: x = [Ba2+] = 2

4[SO ] = 1.52 10–4M The atomic mass of Ba is 137.33 g/mol. The mass of Ba in 1.00 L is: 1.52 10–4M 137.33 g/mol 1.00 L = 0.0209 g 23. Public Health Service standards for drinking water set a maximum of 250 mg/L (2.60 10–

3M) of 24SO because of its cathartic action (it is a laxative). Does natural water that is saturated

http://www.hbcpnetbase.com/)


of 27

with CaSO4 (“gyp” water) as a result or passing through soil containing gypsum, CaSO4.2H2O, meet these standards? What is 2

4SO in such water? Solution First, find the concentration in a saturated solution of CaSO4. Before placing the CaSO4 in water, the concentrations of Ca2+ and 2

4SO are 0. Let x be the change in concentration of Ca2+, which is equal to the concentration of 2

4SO : Ksp = [Ca2+] 2

4[SO ] = 6.1 10–5 x x = x2 = 6.1 10–5

5 = 6.1 10x x = 7.8 10–3M = 2

4[SO ] = [Ca2+] Since this concentration is higher than 2.60 10–3M, “gyp” water does not meet the standards. 24. Perform the following calculations: (a) Calculate [Ag+] in a saturated aqueous solution of AgBr. (b) What will [Ag+] be when enough KBr has been added to make [Br–] = 0.050 M? (c) What will [Br–] be when enough AgNO3 has been added to make [Ag+] = 0.020 M? Solution (a) The dissolution of AgBr produces one Ag+ ion for each Br– ion. Let x equal the change in concentration of Ag+. Initially, the concentrations are 0 for Ag+ and Br–: Ksp = 3.3 10–13 = [Ag+][Br–] = x2 x = [Ag+] = [Br–] = 5.7 10–7M; (b) Ksp = 5.0 10–13 = [Ag+][Br–] = [Ag+]0.050

13+ 115.0 10[Ag ] = = 1.0 10

0.050M

;

(c) Ksp = 5.0 10–13 = [Ag+][Br–] = 0.020[Br–] 13

113.3 10[Br ] = = 1.6 10 0.020

M

25. The solubility product of CaSO42H2O is 6.1 10–5. What mass of this salt will dissolve in1.0 L of 0.010 M 2

4SO ? Solution The amount of CaSO42H2O that dissolves is limited by the presence of a substantial amount of

24SO already in solution from the 0.010 M 4SO . This is a common-ion problem. Let x be the

change in concentration of Ca2+ and of 24SO that dissociates from CaSO4:

2 24 4CaSO ( ) Ca ( ) SO ( )s aq aq

Ksp = [Ca2+] 24[SO ] = 6.1 10–5

Addition of 0.010 M 24SO generated from the complete dissociation of 0.010 M SO4 gives

[x][x + 0.010] = 6.1 10–5. Here, x cannot be neglected in comparison with 0.010 M; the quadratic equation must be used. In standard form: x2 + 0.010x – 6.1 10–5 = 0

4 4 20.01 1 10 2.4 10 0.01 1.8 10 = = 2 2

x


of 27

Only the positive value will give a meaningful answer: x = 4.2 10–3 = [Ca2+] This is also the concentration of CaSO42H2O that has dissolved. The mass of the salt in 1 L is Mass (CaSO42H2O) = 4.2 10–3 mol/L 172.16 g/mol = 0.72 g/L Note that the presence of the common ion, 2

4SO , has caused a decrease in the concentration of Ca2+ that otherwise would be in solution:

5 36.1 10 = 7.8 10 M 26. Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Appendix J for solubility products). (a) TlCl (b) BaF2 (c) Ag2CrO4 (d) CaC2O4•H2O (e) themineral anglesite, PbSO4 Solution (a) Let x be the change in solubility of TlCl in moles per liter upon solution:

[Tl+][Cl–] = Ksp = 1.7 10–4 = x2 x= 1.3 10–2M = [Tl+] = [Cl–] (molar solubility); (b) Let x be the change in solubility of Ba2+ in moles per liter upon solution:

[Ba2+][F–]2 = (x)(2x)2 = Ksp = 2.4 10–5 4x3 = 2.4 10–5 x3 = 6.0 10–6 x= 1.8 10–2M = [Ba2+] [F–] = 2x = 2(1.8 10–2 M) = 3.6 10–2M; (c) Let x be the change in solubility of 2

4CrO in moles per liter upon solution:

[Ag2+] 2

4[CrO ] 2 = (2x)2(x) = Ksp = 9 10–12 4x3 = 9 10–12 x3 = 2.25 10–12


of 27

x = 1.3 10–4M = 24[CrO ]

[Ag+] = 2(1.3 10–4M) = 2.6 10–4M; (d) Let x be the change in solubility of Ca2+ in moles per liter upon solution:

[Ca2+] 2

2 4[C O ] [H2O] = x(x)(55.5) = 55.5x2 = Ksp = 2.27 10–9 The concentration of water in 1 L is 55.5 M. x2 = 3.53 10–10 x = 1.88 10–5 = [Ca2+] = 2

4[CrO ] (e) PbSO4 (Ksp = 1.3 10–8):

2 24 4PbSO ( ) Pb ( ) + SO ( )s aq aq � ��

Let x be the change in solubility of PbSO4 in moles per liter upon solution:

[Pb2+] 2

4[SO ] = Ksp = 1.3 10–8 x2 = 1.3 10–8 x = 1.1 10–4M (molar solubility) 27. Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Appendix J for solubility products): (a) AgI (b) Ag2SO4 (c) Mn(OH)2 (d) Sr(OH)2.8H2O (e) themineral brucite, Mg(OH)2 Solution In each of the following, allow x to be the molar concentration of the ion occurring only once in the formula. (a) Ksp = [Ag+][Cl–] = 1.6 – 10–10 = [x2], [x] = 1.3 – 10–5M, [Ag+] = [I–] = 1.3 10–5M; (b) Ksp =

2 24[Ag ] [SO ] = 1.2 10–5 = [2x]2[x], 4x3 = 1.2 10–5, x = 1.44 10–2M

As there are 2 Ag+ ions for each 24SO ion, [Ag+] = 2.88 10–2M, 2

4[SO ] = 1.44 10–2M; (c) Ksp = [Mn2+]2[OH–]2 = 2 10–13 = [x][2x]2, 4x3 = 2 10–13, x = 3.68 10–5M. Since there are two OH– ions for each Mn2+ ion, multiplication of x by 2 gives 7.36 10–5M. If the value of x is rounded to the correct number of significant figures, [Mn2+] = 3.7 10–5M. [OH–] = 7.4 10–5M. We normally maintain one additional figure in the calculator throughout all calculations before rounding. (d) Ksp = [Sr2+][OH–]2 = 3.2 10–4 = [x][2x]2, 4x3 = 3.2 10–4, x = 4.3 10–2M. Substitution gives [Sr2+] = 4.3 10–2M, [OH–] = 8.6 10–2M;


of 27

(e) Ksp = [Mg2+]2[OH–]2 = 8.9 10–12 = [x][2x]2, 4x3 = 8.9 10–12, x = 1.31 10–4M, 2x = 2.6 10–4. Substitution and taking the correct number of significant figures gives [Mg2+] = 1.3 10–4M, [OH–] = 2.6 10–4M. 28. The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate Ksp for each of the slightly soluble solids indicated: (a) AgBr: [Ag+] = 5.7 10–7M, [Br–] = 5.7 10–7M (b) CaCO3: [Ca2+] = 5.3 10–3M, 2

3[CO ] = 9.0 10–7M (c) PbF2: [Pb2+] = 2.1 10–3M, [F–] = 4.2 10–3M (d) Ag2CrO4: [Ag+] = 5.3 10–5M, 3.2 10–3M (e) InF3: [In3+] = 2.3 10–3M, [F–] = 7.0 10–3M Solution In each case the value of Ksp is found by multiplication of the concentrations raised to the ion’s stoichiometric power. Molar units are not normally shown in the value of K. (a) AgBr: Ksp = [Ag+][Br–] = (5.7 10–7)(5.7 10–7) = 3.2 10–13; (b) CaCO3: Ksp = [Ca+] 2

3[CO ] = (5.3 10–3)(9.0 10–7) = 4.8 10–9; (c) PbF2 = Ksp = [Pb2+][F–]2 = (2.1 10–3)(4.2 10–3)2 = 3.7 10–8; (d) Ag2CrO4: Ksp = 2 2

4[Ag ] [CrO ] = (5.3 10–5)2(3.2 10–3) = 9.0 10–12; (e) InF3: Ksp = [In3+][F–]3 = (2.3 10–3)(7.0 10–3)3 = 7.9 10–10 29. The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate Ksp for each of the slightly soluble solids indicated: (a) TlCl: [Tl+] = 1.21 10–2M, [Cl–] = 1.2 10–2M (b) Ce(IO3)4: [Ce4+] = 1.8 10–4M, 3[IO ] = 2.6 10–13M (c) Gd2(SO4)3: [Gd3+] = 0.132 M, 2

4[SO ] = 0.198 M (d) Ag2SO4: [Ag+] = 2.40 10–2M, 2

4[SO ] = 2.05 10–2M (e) BaSO4: [Ba2+] = 0.500 M, 2

4[SO ] = 4.6 10–8M Solution In each case the value of Ksp is found by multiplication of the concentrations raised to the ion’s stoichiometric power. Molar units are not normally shown in the value of K. (a) TlCl: Ksp= (1.21 10–2)(1.2 10–2) = 1.45 10–4; (b) Ce(IO3)4: Ksp = (1.8 10–4)(2.6 10–13)4 = 8.2 10–55; (c) Gd2(SO4)3: Ksp = (0.132)2(0.198)3 = 1.35 10–4; (d) Ag2SO4: Ksp = (2.40 10–2)2(2.05 10–2) = 1.18 10–5; (e) BaSO4: Ksp = (0.500)(2.16 10–10) = 1.08 10–10 30. Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for Ksp values.) (a) KClO4: [K+] = 0.01 M, 4[ClO ] = 0.01 M (b) K2PtCl6:[K+] = 0.01 M, 2

6[PtCl ] = 0.01 M (c) PbI2: [Pb2+] = 0.003 M, [I–] = 1.3 10–3M (d) Ag2S: [Ag+] = 1 10–10M, [S2–] = 1 10–13M Solution


of 27

Precipitation will occur if the ion concentrations in the reaction quotient exceed the amounts allowed by the solubility product constant. To determine whether this occurs, write the expression for Q, substitute the individual concentrations into the expression, perform the indicated mathematics, and then compare the product to Ksp. (a) KClO4: Q = 4[K ][ClO ] = 0.01 0.01 = 1 10–4, Ksp = 1.05 10–2, Ksp > Q, so the substance does not precipitate;(b) K2PtCl6: Q = 2 2

6[K ] [PtCl ] = (0.01)2 0.01 = 1 10–6, Ksp = 7.48 10–6, Ksp > Q, so the compound does not precipitate;(c) PbI2: Q = [Pb2+][I–]2 = 0.003 (1.3 10–3)2 = 5.07 10–9, Ksp = 1.8 10–8, Ksp > Q, so the compound does not precipitate; (d) Ag2S: Q = [Ag+]2[S2–] = (1 10–10)2 1 10–13 = 1 10–33, Ksp = 1.6 10–49, Ksp < Q, so the compound precipitates 31. Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for Ksp values.) (a) CaCO3: [Ca2+] = 0.003 M, 2

3[CO ] = 0.003 M (b) Co(OH)2: [Co2+] = 0.01 M, [OH–] = 1 10–7M (c) CaHPO4: [Ca2+] = 0.01 M, 2

4[HPO ] = 2 10–6M (d) Pb3(PO4)2: [Pb2+] = 0.01 M, 3

4[PO ] = 1 10–13M Solution (a) 2 2

3 3 3CaCO : CaCO ( ) Ca ( ) CO ( )s aq aq Ksp = [Ca2+] 2

3[CO ] = 8.7 10–9 testKsp against Q = [Ca2+] 2

3[CO ] Q = [Ca2+] 2

3[CO ] = (0.003)(0.003) = 9 10–6 Ksp = 8.7 10–9 < 9 10–6 The ion product does exceed Ksp, so CaCO3 does precipitate. (b) 2 –

2 2Co(OH) : Co(OH) ( ) Co ( ) 2OH ( )s aq aq Ksp = [Co2+][OH–]2 = 2.5 10–16 test Ksp against Q = [Co2+][OH–]2 Q = [Co2+][OH–]2 = (0.01)(1 10–7)2 = 1 10–16 Ksp = 2.5 10–16 > 1 10–16 The ion product does not exceed Ksp, so the compound does not precipitate. (c) CaHPO4: (Ksp = 7 10–7): Q = [Ca2+] 2

4[HPO ] = (0.01)(2 10–6) = 2 10–8 < Ksp The ion product does not exceed Ksp, so compound does not precipitate. (d) Pb3(PO4)2: (Ksp = 1 10–54): Q = [Pb2+]3 3 2

4[PO ] = (0.01)3(1 10–13)2 = 1 10–32 > Ksp The ion product exceeds Ksp, so the compound precipitates. 32. Calculate the concentration of Tl+ when TlCl just begins to precipitate from a solution that is0.0250 M in Cl–. Solution [Tl+][Cl–] = Ksp = 1.7 10–4, [Tl+](0.0250) = 1.7 10–4, [Tl+] = 6.8 10–3M 33. Calculate the concentration of sulfate ion when BaSO4 just begins to precipitate from a solution that is 0.0758 M in Ba2+. Solution


of 27

Precipitation of 24SO will begin when the ion product of the concentration of the 2

4SO and Ba2+ ions exceeds the Ksp of BaSO4. Ksp = 2.3 10–8 = [Ba2+] 2

4[SO ] = (0.0758) 24[SO ]

82 7

42.3 10[SO ] = = 3.03 10

0.0758M

34. Calculate the concentration of Sr2+ when SrF2 starts to precipitate from a solution that is 0.0025 M in F–. Solution Precipitation of SrF2 will begin when the ion product of the ions Sr2+ and F– exceed the Ksp of SrF2:

22SrF ( ) Sr ( ) 2F ( )s aq aq � ��

Ksp = [Sr2+][F–]2 = 3.7 10–12 The maximum concentration of Sr2+ that will remain in solution is:

122 7

23.7 10[Sr ] 5.9 10 (0.0025)

M

35. Calculate the concentration of 34PO when Ag3PO4 starts to precipitate from a solution that

is 0.0125 M in Ag+. Solution Precipitation of Ag3PO4 will begin when the ion product of the concentrations of the Ag+ and

34PO ions exceeds Ksp:

33 4 4Ag PO ( ) 3Ag ( ) PO ( )s aq aq

Ksp = 1.8 10–18 = [Ag+]3 34[PO ] = (0.0125)3 3

4[PO ] 18

3 134 3

1.08 10[PO ] = = 9.2 10 (0.0125)

M

36. Calculate the concentration of F– required to begin precipitation of CaF2 in a solution that is 0.010 M in Ca2+. Solution

22CaF ( ) Ca ( ) 2F ( )s aq aq � ��

[Ca2+][F–]2 = Ksp = 4.0 10–11 (0.010)[F–]2 = 4.0 10–11 [F–]2 = 4.0 10–9 [F–] = 6.3 10–5 37. Calculate the concentration of Ag+ required to begin precipitation of Ag2CO3 in a solution that is 2.50 10–6M in 2

3CO . Solution

22 3 3Ag CO ( ) 2Ag ( ) CO ( )s aq aq

[Ag+]2 23[CO ] = Ksp = 8.1 10–12

[Ag+]2(2.5 10–6) = 8.1 10–12 12

+ 2 66

8.1 10[Ag ] = = 3.28 102.50 10


of 27

[Ag+] = 1.8 10–3M 38. What [Ag+] is required to reduce 2

3[CO ] to 8.2 10–4Mby precipitation of Ag2CO3? Solution Ksp = 8.1 10–12 = [Ag+]2(8.2 10–4)

122 9

4

8.1 10[Ag ] 9.9 108.2 10

[Ag+] = 9.9 10–5M 39. What [F–] is required to reduce [Ca2+] to 1.0 10–4M by precipitation of CaF2? Solution In the Ksp expression, substitute the concentration of Ca2+ and solve for [F–]. Ksp = 4.0 10–11 = [Ca2+][F–]2 = (1.0 10–4)[F–]2

112 7

4

4.0 10[F ] = = 4.0 101.0 10

[F–] = 6.3 10–4 40. A volume of 0.800 L of a 2 10–4-M Ba(NO3)2 solution is added to 0.200 L of 5 10–4M Li2SO4. Does BaSO4 precipitate? Explain your answer. Solution Calculate the molar concentration of the Ba2+ and 2

4SO ions in the new total volume as if no precipitation occurred. Then use the Ksp expression to determine whether these concentrations exceed Ksp. If so, BaSO4 precipitates. 2 10–4M 0.800 L = [Ba2+] 1.00 L [Ba2+] = 1.6 10–4M 5 10–4M 0.200 L = 2

4[SO ] 1.00 L 2

4[SO ] = 1.0 10–4M Ksp = 2.3 10–8 = 2 2

4[Ba ][SO ] 2 2

4[Ba ][SO ] = (1.06 10–4)(1.0 10–4) = 1.6 10–8 As the latter value is smaller than 2.3 10–8, a precipitate will not form. 41. Perform these calculations for nickel(II) carbonate. (a) With what volume of water must a precipitate containing NiCO3 be washed to dissolve 0.100 g of this compound? Assume that the wash water becomes saturated with NiCO3 (Ksp = 1.36 10–7). (b) If the NiCO3 were a contaminant in a sample of CoCO3 (Ksp = 1.0 10–12), what mass of CoCO3 would have been lost? Keep in mind that both NiCO3 and CoCO3 dissolve in the same solution. Solution (a) Calculate the molar solubility. Then calculate the number of grams per liter. Ksp = 1.4 10–7 = [Ni2+] 2

3[CO ] Before placement of the sample into water, the concentration of the ions is 0. Let x be the change in concentration of the two ions formed. The total concentration of each is thus 0 + x = x. Thus: x2 = 1.4 10–7 x = [Ni2+] = 2

3[CO ] = 3.742 10–4M The molar mass of NiCO3 is 118.71 g/mol. Thus: Concentration of NiCO3 = 118.71 g/mol 3.742 10–4 mol/L = 0.0444 g/L


of 27

To contain 0.1 g, 10.1 g = 2.25 L

0.0444 g L ;

(b) During the process of removal of NiCO3, some CoCO3 would be lost. The 23[CO ] is

controlled by the amount found in part (a). From the solubility product for CoCO3: Ksp = 1.0 10–12 = [Co2+][3.742 × 10–4] [Co2+] = 2.67 10–9M The molar mass of CoCO3 is 118.94 g/mol. Thus: mass CoCO3 in 2.28 L = 118.94 g/mol 2.67 10–9 mol/L 2.28 L = 7.2 10–7 g 42. Iron concentrations greater than 5.4 10–6M in water used for laundry purposes can cause staining. What [OH–] is required to reduce [Fe2+] to this level by precipitation of Fe(OH)2? Solution [Fe2+][OH–]2 = Ksp = 1.8 10–15 (5.4 × 10–6)[OH–]2 = 1.8 10–15 [OH–]2 = 3.3 10–10 [OH–] = 1.8 10–5M 43. A solution is 0.010 M in both Cu2+ and Cd2+.What percentage of Cd2+ remains in the solution when 99.9% of the Cu2+ has been precipitated as CuS by adding sulfide? Solution When 99.9% of Cu2+ has precipitated as CuS, then 0.1% remains in solution. 0.1100

0.010 mol/L = 1 10–5M = [Cu2+]

[Cu2+][S2–] = Ksp = 8.5 10–45 (1 10–5)[S2–] = 8.5 10–45 [S2–] = 8.5 10–40M [Cd2+][S2–]Ksp = 1.0 10–28 [Cd2+](7 10–37) = 1.0 10–28 [Cd2+] = 1.2 1011M Thus [Cd2+] can increase to 1.2 1011 M before precipitation begins. [Cd2+] is only 0.010 M, so 100% of it is dissolved. 44. A solution is 0.15 M in both Pb2+ and Ag+. If Cl– is added to this solution, what is [Ag+] when PbCl2 begins to precipitate? Solution Lead and silver both form slightly soluble chlorides. The ion with less molar solubility will precipitate first. This precipitation will continue until the chloride ion increases sufficiently to exceed the Ksp of the other chloride. Since both Ag+ and Pb2+ must exist simultaneously in solution, both Ksp equilibria must be satisfied:

2 52 sp

10sp

PbCl ( ) Pb ( ) 2Cl ( ) ( 1.6 10 )

AgCl( ) Ag ( ) Cl ( ) ( 1.6 10 )

s aq aq K

s aq aq K

� ��

� ��

First, calculate the maximum chloride in concentration that can be present without causing the precipitation of Pb2+. Given that [Pb2+] equals 0.15 M, the [Cl–] maximum is: [Pb2+][Cl–]2 = 1.7 10–5

52 41.6 10[Cl ] 1.07 10

0.15

[Cl–] = 1.03 10–2


of 27

Precipitation of AgCl will continue until the chloride ion concentration equals 1.06 10–2M. At this point the [Ag+] is controlled by the Ksp of AgCl. Substituting this [Cl–] into the Ksp of AgCl gives: [Ag+][1.06 10–2] = 1.6 10–10

108

2

1.6 10[Ag ] 1.6 10 1.03 10

M

45. What reagent might be used to separate the ions in each of the following mixtures, which are 0.1 M with respect to each ion? In some cases it may be necessary to control the pH. (Hint: Consider the Ksp values given in Appendix J.) (a) 2

2Hg and Cu2+ (b) 2–

4SO and Cl– (c) Hg2+ and Co2+ (d) Zn2+ and Sr2+ (e) Ba2+ and Mg2+ (f) 2

3CO and OH– Solution To compare ions of the same oxidation state, look for compounds with a common counter ion that have very different Ksp values, one of which has a relatively large Ksp—that is, a compound that is somewhat soluble. (a) 2

2Hg and Cu2+: Add 24SO . CuSO4 is soluble (see Appendix J), but Ksp for Hg2SO4 is 7.4

10–7. When only 0.1% 22Hg remains in solution:

2 42

0.1%[Hg ] = 0.10 = 1 10 100%

M

and 7

sp2 34 2 4

2

7.4 10[SO ] = = 7.4 10 [Hg ] 1 10

KM

;

(b) 24SO and Cl–: Add Ba2+. BaCl2 is soluble (see the section on catalysis), but Ksp for BaSO4 is

2.3 10–8. When only 0.1% 24SO remains in solution, 2

4[SO ] = 1 10–4M and 8

2 44

2.3 10[Ba ] = = 2.3 10 1 10

M

;

(c) Hg2+ and Co2+: Add S2–: For the least soluble form of CoS, Ksp = 3 10–26 and for HgS, Ksp = 1.6 10–54. CoS will not begin to precipitate until: [Co2+][S2–] = Ksp = 3 10–26 (0.10)[S2–] = 3 10–26 [S2–] = 3 10–25 At that concentration: [Hg2+](3 10–25) = 1.6 10–54 [Hg2+] = 5.3 10–30 M That is, it is virtually 100% precipitated. For a saturated (0.10 M) H2S solution, the corresponding +

3[H O ] is:


of 27

2723 a2 25

[H S] (0.10)[H O ] = = = 8.9 10[S ] (3 10 )

K

+3[H O ] = 0.0030 M

A solution more basic than this would supply enough S2– for CoS to precipitate. (d) Zn2+ and Sr2+: Add OH– until [OH–] = 0.050 M. For Sr(OH)28H2O, Ksp = 3.2 10–4. For Zn(OH)2, Ksp = 4.5 10–11. When Zn2+ is 99.9% precipitated, then [Zn2+] = 1 10–4M and

11sp2 72 4

4.5 10[OH ] = = = 4.5 10[Zn ] 1 10

K

[OH–] = 7 10–4M When Sr(OH)28H2O just begins to precipitate:

4sp2 32

3.2 10[OH ] = = = 3.2 10[Sr ] 0.10

K

[OH–] = 0.057 M If [OH–] is maintained less than 0.056 M, then Zn2+ will precipitate and Sr2+ will not. (e) Ba2+ and Mg2+: Add 2

4SO . MgSO4 is soluble and BaSO4 is not (Ksp = 2.3 10–8). (f) 2

3CO and OH–: Add Ba2+. For Ba(OH)2, 8H2O, Ksp = 5.0 10–3; for BaCO3, Ksp = 1.6 10–

9. When 99.9% of 23CO has been precipitated 2

3[CO ] = 1 10–4M and 9

sp2+ 52 4

3

1.6 10[Ba ] = = = 1.6 10 [CO ] 1 10

KM

Ba(OH)28H2O begins to precipitate when: 3

sp2+2 2

5.0 10[Ba ] = = = 0.50 [OH ] (0.10)

KM

As long as [Ba2+] is maintained at less than 0.50 M, BaCO3 precipitates and Ba(OH)28H2O does not. 46. A solution contains 1.0 10–5 mol of KBr and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. Which forms first, solid AgBr or solid AgCl? Solution Compare the concentration of Ag+ as determined from the two solubility product expressions. The one requiring the smaller [Ag+] will precipitate first. For AgCl: Ksp = 1.6 10–10 = [Ag+][Cl–]

–10+ 91.6 10[Ag ] = = 1.6 10

0.10M

For AgI: Ksp = 1.5 10–16 = [Ag+][I–] –16

+ 14–2

1.5 10[Ag ] = = 1.5 10 1.0 10

M

As the value of [Ag+] is smaller for AgI, AgI will precipitate first. 47. A solution contains 1.0 10–2 mol of KI and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. Which forms first, solid AgI or solid AgCl? Solution Compare the concentration of Ag+ as determined from the two solubility product expressions. The one requiring the smaller [Ag+] will precipitate first.


of 27

For AgCl: Ksp = 1.6 10–10 = [Ag+][Cl–] 10

+ 91.6 10[Ag ] = = 1.6 10 [0.10]

M

For AgI: Ksp = 1.5 10–16 = [Ag+][I–] 16

+ 92

1.5 10[Ag ] = = 1.5 10 1.0 10

M

As the value of [Ag+] is smaller for AgI, AgI will precipitate first. 48. The calcium ions in human blood serum are necessary for coagulation (Figure 15.5). Potassium oxalate, K2C2O4, is used as an anticoagulant when a blood sample is drawn for laboratory tests because it removes the calcium as a precipitate of CaC2O4.H2O. It is necessary to remove all but 1.0% of the Ca2+ in serum in order to prevent coagulation. If normal blood serum with a buffered pH of 7.40 contains 9.5 mg of Ca2+ per 100 mL of serum, what mass of K2C2O4 is required to prevent the coagulation of a 10mL blood sample that is 55% serum by volume? (All volumes are accurate to two significant figures. Note that the volume of serum in a 10-mL blood sample is 5.5 mL. Assume that the Ksp value for CaC2O4 in serum is the same as in water.) Solution First, find the number of moles of Ca2+ present that must be precipitated. If 9.5 mg of Ca2+ is present in 100 mL of serum, 5.5 mL contains:

29.5 mg 5.5 mL 0.52 mg Ca100 mL

4 25 2+

15.2 10 g Ca = 1.297 10 mol Ca

40.078 g mol

Next, 99% of this amount—1.284 10–5 mol—must be removed. This process requires 1.284 10–5 mol of 2

2 4C O plus the amount of 22 4C O required to maintain the Ksp equilibrium. The

number of moles of Ca2+ remaining in solution is 1.2947 10–5 – 1.284 10–5 = 1.3 10–7 mol Ca2+. These ions are present in a 10-mL blood sample, so the molarity is:

75 2+1.3 10 mol = 1.3 10 Ca

0.010 LM

From the Ksp expression: 8

2 3 12 4 5

1.96 10[C O ] = = 1.51 10 mol L1.3 10

The moles of 22 4C O for this latter purpose total:

1.51 10–3M 0.01 L = 1.51 10–5 mol Thus the total number of moles of 2

2 4C O required is: 1.284 10–5 mol + 1.51 10–5 mol = 2.79 10–5 mol mass (K2C2O4) = 2.79 10–5 mol 166.2162 g/mol = 4.6 10–3 g 49. About 50% of urinary calculi (kidney stones) consist of calcium phosphate, Ca3(PO4)2. The normal mid range calcium content excreted in the urine is 0.10 g of Ca2+ per day. The normal mid range amount of urine passed may be taken as 1.4 L per day. What is the maximum concentration of phosphate ion that urine can contain before a calculus begins to form? Solution The dissolution of Ca3(PO4)2 yields:


of 27

2+ 33 4 2 4Ca (PO ) ( ) 3Ca ( ) + 2PO ( )s aq aq� ��

Given the concentration of Ca2+ in solution, the maximum 34[PO ] can be calculated by using the

Ksp expression for Ca3(PO4)2: Ksp = 1.3 10–32 = [Ca2+]3 3 2

4[PO ]

2+ 3urine

1 mol0.10 g40.08 g

[Ca ] = = 1.8 10 1.4 L

M

323 2 24

4 3 31.3 10[PO ] = = 2.2 10

(1.8 10 )

3 2 124[PO ] = 1.5 10 M

50. The pH of normal urine is 6.30, and the total phosphate concentration ( 34[PO ] + 2

4[HPO ] + 2 4[H PO ] + [H3PO4]) is 0.020 M. What is the minimum concentration of Ca2+ necessary to induce kidney stone formation? (See Exercise 49 for additional information.) Solution The concentration of Ca2+ depends on the concentration of 3

4PO in solution. 34[PO ] is

dependent on pH and the H3PO4 equilibrium. Because the total phosphate concentration is 0.020 M, 3

4[PO ] must be calculated from the three expressions involving H3PO4. These expressions, solved in terms of 3

4[PO ] , yield a ratio of 34[PO ] of the other components:

33 4 2 3 2 4 1

2 82 4 2 3 4 2

2 3 134 2 3 4 3

H PO ( ) H O( ) H O ( ) H PO ( ) ( 7.5 10 )

H PO ( ) H O( ) H O ( ) HPO ( ) ( 6.2 10 )

HPO ( ) H O( ) H O ( ) PO ( ) ( 4.2 10 )

aq l aq aq K

aq l aq aq K

aq l aq aq K

� ��

� ��

� ��

At a pH of 6.3, the7

3[H O ] = 5.01 10 . For equation (3): 3

133 42

47

2 3 6 34 4 413

[H O ][PO ] 4.2 10[HPO ]

5.01 10[HPO ] [PO ] 1.19 10 [PO ]3.6 10

For equation (2): 2

83 4

2 4

[H O ][HPO ] 6.2 10[H PO ]

72 2 6 33

2 4 4 4 48 8

6 34

[H O ] 5.01 10[H PO ] [HPO ] [HPO ] = 8.08(1.19 10 )[PO ] 6.2 10 6.2 10

9.62 10 [PO ]

For equation (1):


of 27

33 2 4

3 4

75 6 3 3

3 4 2 4 4 43

3 24 4 2 4 3 4

6 6

[H O ][H PO ] 7.5 10[H PO ]

5.01 10[H PO ] [H PO ] 6.7 10 (9.62 10 )[PO ] 645[PO ]7.5 10

[PO ]:[HPO ]:[H PO ]:[H PO ]

1:1.19 10 :9.62 10 : 645

Adding the four numerical terms above gives 1.08 107. The fraction of 34PO in solution is:

384

3 2 74 4 2 4 3 4

[PO ] 1 9.26 10[PO ] [HPO ] [H PO ] [H PO ] [1.08 10 ]

The concentration of 34PO in the urine at pH 6.3 is:

34[PO ] = 9.26 10–8[0.020] = 1.85 10–9

Use the Ksp of Ca3(PO4)2 to calculate the minimum [Ca2+] present. For the reaction: 2 3

3 4 2 4Ca (PO ) ( ) 3Ca ( ) 2PO ( ),s aq aq � �� 2 3 3 2

sp 4

322 3 15

3 2 9 24

2 8 1/3 5

[Ca ] [PO ]

1.3 10[Ca ] 3.80 10 [PO ] (1.85 10 )

[Ca ] (3.96 10 ) 1.56 10

sp

KK

M

M

51. Magnesium metal (a component of alloys used in aircraft and a reducing agent used in the production of uranium, titanium, and other active metals) is isolated from sea water by the following sequence of reactions:

2 22 2Mg ( ) Ca(OH) ( ) Mg(OH) ( ) Ca ( )aq aq s aq

2 2 2Mg(OH) ( ) 2HCl( ) MgCl ( ) 2H O( )s aq s l electrolysis

2 2MgCl ( ) Mg( ) Cl ( )l s g Sea water has a density of 1.026 g/cm3 and contains 1272 parts per million of magnesium as Mg2+(aq) by mass. What mass, in kilograms, of Ca(OH)2 is required to precipitate 99.9% of the magnesium in 1.00 103 L of sea water? Solution Calculate the amount of Mg2+ present in sea water; then use Ksp to calculate the amount of Ca(OH)2 required to precipitate the magnesium. mass Mg = 1.00 103 L 1000 cm3/L 1.026 g/cm3 1272 ppm 10–6 ppm–1 = 1.305 103 g The concentration is 1.305 g/L. If 99.9% is to be recovered 0.999 1.305 g/L = 1.304 g/L will be obtained. The molar concentration is:

1

11.304 g L 0.05365

24.305 g molM

As the Ca(OH)2 reacts with Mg2+ on a 1:1 mol basis, the amount of Ca(OH)2 required to precipitate 99.9% of the Mg2+ in 1 L is: 0.05365 M 74.09 g/mol Ca(OH)2 = 3.97 g/L


of 27

For treatment of 1000 L, 1000 L 3.97 g/L = 3.97 103 g = 3.97 kg. However, additional [OH–

] must be added to maintain the equilibrium: 2 12

2 spMg(OH) ( ) Mg ( ) 2OH ( ) ( 8.9 10 )s aq aq K � �� When the initial 1.035 g/L is reduced to 0.1% of the original, the molarity is calculated as:

15

10.001 1.305 g L 5.369 10

24.305 g molM

The added amount of OH– required is found from the solubility product: [Mg2+][OH–]2 = (5.369 10–5)[OH–]2 = 8.9 10–12 = Ksp [OH–] = 4.07 10–4

Thus an additional 12 4.07 10–4 mol (2.65 10–4 mol) of Ca(OH)2 per liter is required to

supply the OH–. For 1000 L: 4 1 3

2 22

74.0946 gmass Ca(OH) 2.04 10 mol Ca(OH) L 1.00 10 L 15 gmol Ca(OH)

The total Ca(OH)2 required is 3.97 kg + 0.015 kg = 3.99 kg. 52. Hydrogen sulfide is bubbled into a solution that is 0.10 M in both Pb2+ and Fe2+ and 0.30 M in HCl. After the solution has come to equilibrium it is saturated with H2S ([H2S] = 0.10 M). What concentrations of Pb2+ and Fe2+ remain in the solution? For a saturated solution of H2S we can use the equilibrium:

2– –272 2 3H S( ) 2H O( ) 2H O ( ) S ( ) 8.9 10aq l aq aq K � ��

(Hint: The 3[H O ] changes as metal sulfides precipitate.) Solution Ksp = [Fe2+][S2–] = 3.7 10–19; Ksp = [Pb2+][S2–] = 7 10–29

Initially, 2 2 2 2

273

2

[H O ] [S ] (0.30) [S ] = = = 8.9 10[H S] (0.10)

K

[S2–] = 9.9 10–27M The concentrations of Fe2+ and Pb2+ are calculated as follows: Ksp = [Fe+][S2–] = [Fe2+](9.9 10–27) = 3.7 10–19 [Fe3+] = 3.4 107 M Ksp = [Pb2+][S2–] = [Pb2+](9.9 10–27) = 7 10–29 [Pb2+] = 71M. Thus, all the iron and lead ions originally present will remain in solution, 0.10 M for each. 53. Perform the following calculations involving concentrations of iodate ions: (a) The iodate ion concentration of a saturated solution of La(IO3)3 was found to be 3.1 10–3 mol/L. Find the Ksp. (b) Find the concentration of iodate ions in a saturated solution of Cu(IO3)2 (Ksp = 7.4 10–8). Solution

(a) Ksp = [La3+] 33[IO ] = 31 3.1 10

3

(3.1 10–3)3 = (0.0010)(3.0 10–8) = 3.1 10–11;

(b) Ksp = [Cu2+] 23[IO ] = x(2x)2 = 7.4 10–8

4x3 = 7.4 10–8 x3 = 1.85 10–8


of 27

x = 2.64 10–3 [Cu2+] = 2.6 10–3

3[IO ] = 2x = 5.3 10–3 54. Calculate the molar solubility of AgBr in 0.035 M NaBr (Ksp = 5 10–13). Solution

(x)(0.035 M + x) = 5 10–13 If we ignore the +x, then x is equal to 1.4 10–11M. This is well below the 5% rule, so we can ignore the +x. 1.4 10–11M 55. How many grams of Pb(OH)2 will dissolve in 500 mL of a 0.050-M PbCl2 solution (Ksp = 1.2 10–15)? Solution

(2x)(0.050 M + x) = 1.2 10–15 If we ignore the +x, then x is equal to 1.2 10–14M. This is well below the 5% rule, so we can ignore the +x:

0.500 L141.2 10 mol

1 L

241.22 g 1 mol

12 = 1.4 10 g

1.8 10–5 g Pb(OH)2 56. Use the simulation (http://employees.oneonta.edu/viningwj/sims/common_ion_effect_s1.html) from the earlier Link to Learning to complete the following exercise: Using 0.01 g CaF2, give the Ksp values found in a 0.2-M solution of each of the salts. Discuss why the values change as you change soluble salts. Solution After going through the simulation, the solubility of CaF2, is 7.59 10–8 g/L when placed in 0.2 M of Na F, KF, and Ca(NO3)2. When placed in NaNO3, the solubility is 1.66 10–2 g/L. The solubility is highest in NaNO3 because there is no common ion between CaF2 and NaNO3. When a common ion is present (as for the other three solutions, either F– or Ca2+), the reaction is going to shift towards the reactant (the solid) according to Le Châtelier’s principle. 57. How many grams of Milk of Magnesia, Mg(OH)2 (s) (58.3 g/mol), would be soluble in 200 mL of water. Ksp = 8.9 10–12. Include the ionic reaction and the expression for Ksp in your answer. What is the pH? (Kw = 1 10–14 = 3[H O ] [OH–]) Solution

http://employees.oneonta.edu/viningwj/sims/common_ion_effect_s1.html)


of 27

2 2 22 spMg(OH) ( ) Mg 2OH [Mg ][OH ]s K � ��

8.9 10–12 = (x)(2x)2 = 4x3 x = 1.31 10–4M = [Mg2+]

432 2

222

1 mol Mg(OH) 58.3 g Mg(OH)1.31 10 mol 0.200 L × = 1.23 10 g Mg(OH)L 1 mol Mg 1 mol Mg(OH)

58. Two hypothetical salts, LM2 and LQ, have the same molar solubility in H2O. If Ksp for LM2 is 3.20 10–5, what is the Ksp value for LQ? Solution

2 2 22 spLM (s) L 2M [L ][M ]K � ��

3.20 10–5 = (x)(2x)2 = 4x3 x = 0.020 M = [L2+]

2 2 2 2spLQ( ) L Q [L ][Q ]s K � ��

Same molar solubility for LM2 as for LQ: Ksp = (0.020 M)(0.020 M) = 0.00040 59. The carbonate ion concentration is gradually increased in a solution containing divalent cations of magnesium, calcium, strontium, barium, and manganese. Which of the following carbonates will form first? Which of the following will form last? Explain. (a) 5

3 spMgCO 1 10K

(b) 93 spCaCO 8.7 10K

(c) 103 spSrCO 7 10K

(d) 93 spBaCO 1.6 10K

(e) 113 spMnCO 8.8 10K

Solution MnCO3 will form first, since it has the smallest Ksp value it is the least soluble. MgCO3 will be the last to precipitate, it has the largest Ksp value. 60. How many grams of Zn(CN)2(s) (117.44 g/mol) would be soluble in 100 mL of H2O?Include the balanced reaction and the expression for Ksp in your answer. The Ksp value for Zn(CN)2(s) is 3.0 10–16. Solution

22 2sp2

Zn CN ( ) Zn 2CN Zn CNs K � �� 3.0 10–16 = (x)(2x)2 = 4x3 x = 4.22 10–6M

6 252 2

222

1 mol Zn(CN) 117.44 g ZN(CN)4.22 10 mol Zn 0.100 L = 4.96 10 g Zn(CN)L 1 mol Zn 1 mol Zn(CN)

61. Even though Ca(OH)2 is an inexpensive base, its limited solubility restricts its use. What is the pH of a saturated solution of Ca(OH)2? Solution If we let be the change in concentration of Ca2+, then: 2

2Ca ( ) 2OH Ca(OH) ( )aq s � ��


of 27

Ksp = [Ca2+][OH–]2 = (x)(2x)2 = 1.3 10–6 4x3 = 1.3 10–6 x3 = 3.25 10–7 x = 0.00688 = [Ca2+] [OH–] = 0.0138 M, pOH = 1.86, pH = 12.14 This resource file is copyright 2019, Rice University. All Rights Reserved.

OpenStax Chemistry 2e 15.2: Lewis Acids and Bases

Page 1 of 10


15.2: Lewis Acids and Bases 62. Under what circumstances, if any, does a sample of solid AgCl completely dissolve in pure water? Solution when the amount of solid is so small that a saturated solution is not produced 63. Explain why the addition of NH3 or HNO3 to a saturated solution of Ag2CO3 in contact with solid Ag2CO3 increases the solubility of the solid. Solution NH3 forms a complex ion with Ag+, reducing its concentration. HNO3 reacts with 2

3CO , reducing its concentration. 64. Calculate the cadmium ion concentration, [Cd2+], in a solution prepared by mixing 0.100 L of 0.0100 M Cd(NO3)2 with 1.150 L of 0.100 NH3(aq). Solution Cadmium ions associate with ammonia molecules in solution to form the complex ion

2+3 4[Cd(NH ) ] , which is defined by the following equilibrium:

22 7

3 3 f4Cd ( ) 4NH ( ) Cd NH ( ) 1.3 10aq aq aq K

The formation of the complex ion requires 4 mol of NH3 for each mol of Cd2+. First, calculate the initial amounts of Cd2+ and of NH3 available for association:

12+ 3(0.100 L)(0.0100 mol L )[Cd ] = = 4.00 10

0.250 LM

12

3(0.150 L)(0.100 mol L )[NH ] = = 6.00 10

0.250 LM

For the reaction, 4.00 10–3mol/L of Cd2+ would require 4(4.00 10–3 mol/L) of NH3 or a 1.6 10–2-M solution. Due to the large value of Kf and the substantial excess of NH3, it can be assumed that the reaction goes to completion with only a small amount of the complex dissociating to form the ions. After reaction, concentrations of the species in the solution are [NH3] = 6.00 10–2 mol/L – 1.6 10–2 mol L–1 = 4.4 10–2 M Let x be the change in concentration of [Cd2+]:

2

6 3 4f 2 4

3

[Cd(NH ) ] = 4.0 10 = [Cd ][NH ]

K

37

2 4(4.00 10 )1.3 10 =

( )(4.4 10 4 )x

x x

As x is expected to be about the same size as the number from which it is subtracted, the entire expression must be expanded and solved, in this case, by successive approximations where


Page 2 of 10

substitution of values for x into the equation continues until the remainder is judged small enough. This is a slightly different method than used in most problems. We have: 1.3 107x (4.4 10–2 + 4x)4 = 4.00 10–3 – x 1.3 107x (3.75 10–6 + 1.36 10–3x + 0.186x2 + 11.264x3 + 256x4) = 4.00 10–3 16x + 5440x2 + 7.44 105x3 + 4.51 107x4 + 1.024 109x5 = 4.00 10–3 Substitution of different values x will give a number to be compared with 4.00 10–3. Using 8 10–5 gives 4.01 10–3, this is a good fit. Thus 8 105 is close enough to the true value of x to make the difference equal to zero. The decision to drop 4x compared with 4.4 x 10–2 is justified. 65. Explain why addition of NH3 or HNO3 to a saturated solution of Cu(OH)2 in contact with solid Cu(OH)2 increases the solubility of the solid. Solution NH3 forms a complex ion with Cu2+, reducing its concentration and allowing more Cu2+ ions to dissolve from the solid phase. 3HNO reacts with OH–, thereby reducing the OH– concentration and allowing more solid to dissolve. 66. Sometimes equilibria for complex ions are described in terms of dissociation constants, Kd. For the complex ion 3–

6AlF the dissociation reaction is:

3– 36AlF Al 6F � �� and

3 624

d 36

[Al ][F ] = = 2 10[AlF ]

K

Calculate the value of the formation constant, Kf, for 3–6AlF .

Solution For the formation reaction:

3 36

3236

f 3 6 24d

Al ( ) 6F ( ) AlF ( )

[AlF ] 1 1 5 10[Al ][F ] 2 10

aq aq aq

KK

� ��

67. Using the value of the formation constant for the complex ion 23 6Co(NH ) , calculate the

dissociation constant. Solution The formation constant is the reciprocal of the dissociation constant:

253 6

f 2 63 d

[Co(NH ) ] 1 = = 1.3 10 = [Co ][NH ]

KK

Kd = 7.7 10–6 68. Using the dissociation constant, Kd = 7.8 10–18, calculate the equilibrium concentrations of Cd2+ and CN– in a 0.250-M solution of 2

4Cd(CN) . Solution

2 4

18d 2

4

[Cd ][CN ] (4 ) 7.8 10 [Cd(CN) ] 0.250

x xKx


Page 3 of 10

Assume that x is small when compared with 0.250 M. 256x5 = 0.250 7.8 10–18 x5 = 7.617 10–21 x = [Cd2+] = 9.5 10–5 M 4x = [CN–] = 3.8 10–4 M 69. Using the dissociation constant, Kd = 3.4 10–15, calculate the equilibrium concentrations of Zn2+ and OH– in a 0.0465-M solution of 2

4Zn(OH) . Solution

2 4 4

d 24

[Zn ][OH ] (4 ) = = 3.4 10 15 = [Zn(OH) ] 0.0465

x xKx

Assume that x is small when compared with 0.0465 M. 256x5 = 0.0465 3.4 10–15 x5 = 6.176 10–19 x = [Zn2+] = 2.3 10–4 M 4x = [OH–] = 9.1 10–4 M 70. Using the dissociation constant, Kd = 2.2 10–34, calculate the equilibrium concentrations of Co3+ and NH3 in a 0.500-M solution of 3

3 6Co(NH ) . Solution

2 6 6

343d 3

3 6

[Co ][NH ] (6 ) 2.2 10[Co(NH ) ] 0.500

x xKx

Assume that x is small when compared with 0.500 M. 4.67 104x7 = 0.500 2.2 10–34 x7 = 2.358 10–39 x = [Co3+] = 3.0 10–6 M 6x = [NH3] = 1.8 10–5 M 71. Using the dissociation constant, Kd = 1 10–44, calculate the equilibrium concentrations of Fe3+ and CN– in a 0.333 M solution of 3

6Fe(CN) . Solution


Page 4 of 10

3 6 644

d 36

[Fe ][CN ] (6 ) = = = 1 10[Fe(CN) ] 0.333

x xKx

Assume that x is small when compared with 0.500 M. 4.67 104x7 = 3.33 10–45 x7 = 7.137 10–50 x = [Fe3+] = 1 10–7 M 6x = [CN–] = 6 10–7 M 72. Calculate the mass of potassium cyanide ion that must be added to 100 mL of solution to dissolve 2.0 10–2 mol of silver cyanide, AgCN. Solution Because Ksp is small and Kf is large, most of the Ag+ is used to form 2Ag(CN) ; that is:

21

2

[Ag ] < [Ag(CN) ]

[Ag(CN) ] 2.0 10 M

The CN– from the dissolution and the added CN– exist as CN–and 2Ag(CN) . Let x be the change in concentration upon addition of CN–. Its initial concentration is approximately 0. [CN–] + 2 2[Ag(CN) ] = 2 10–1 + x Because Ksp is small and Kf is large, most of the CN– is used to form 2[Ag(CN) ] ; that is:

2[CN ] < 2[Ag(CN) ] . 1

22[Ag(CN) ] 2.0 10 x 2(2.0 10–1) – 2.0 10–1 = x 2.0 10–1 M L = mol CN– added The solution has a volume of 100 mL. 2 10–1 mol/L 0.100 L = 2 10–2 mol mass KCN = 2.0 10–2 mol KCN 65.120 g/mol = 1.3 g 73. Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve 3.0 10–3 mol of silver bromide. Solution Two simultaneous equilibria are involved:

–3 3 2AgBr( ) 2NH ( ) Ag(NH ) ( ) Br ( )s aq aq aq

(1) +

73 2f + 2

3

[Ag(NH ) ] = = 1.7 10[Ag ][NH ]

K

(2) –AgBr( ) Ag ( ) BrH ( )s aq aq Ksp = [Ag+][Br–] = 5 10–13 (3) Express [Ag+] in terms of Br– and Ksp, and then substitute into equation (2).

13+ 5 × 10[Ag ] =

[Br ]

73 213 2

3

[Ag(NH ) ][Br ] = 1.7 105 10 [NH ]

(4) When 3.0 10–3 mol of silver bromide dissolves, [Br–] = 3.0 10–3 M, and +3 2[Ag(NH ) ] =

[Br–]. Substitution into equation (4) gives, after rearrangement:


Page 5 of 10

3 22 23 2

3 13 7 6

[Ag(NH ) ][Br ] (3.0 10 )[NH ] = = = 1.06 (5 10 )(1.7 10 ) 8.5 10

M

[NH3] = 1.0 M 74. A roll of 35-mm black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of Na2S2O3.5H2O (sodium thiosulfate pentahydrate or hypo) in 1.0 L of developer is required to dissolve the AgBr as 3

2 3 2Ag(S O ) (Kf = 4.7 1013)? Solution The reaction is governed by two equilibria, both of which must be satisfied:

13sp

2 3 132 3 2 3 2 f

AgBr( ) Ag ( ) Br ( ) 5 10

Ag ( ) 2S O ( ) Ag(S O ) ( ) 4.7 10

s aq aq K

aq aq aq K

� ��

� ��

The overall equilibrium is obtained by adding the two equations and multiplying their Ks: 3

2 32 2

2 3

[Ag(S O ) ][Br ] 23.5[S O ]

If all Ag is to be dissolved, the concentration of the complex is the molar concentration of AgBr. formula mass (AgBr) = 187.772 g/mol

31

0.27 g AgBrmoles present 1.438 10 mol187.772 g mol

Let x be the change in concentration of 22 3S O :

3 3

2

(1.438 10 )(1.438 10 ) 23.5x

x2 = 8.799 10–8 4 2

2 3 2.97 10 [S O ]x M The formula mass of Na2S2O3·5H2O is 248.13 g/mol. The total 2

2 3[S O ] needed is: 2(1.438 10–3) + 2.97 10–4 = 3.173 10–3 mol g(hypo) = 3.173 10–3 mol 248.13 g/mol = 0.79 g 75. We have seen an introductory definition of an acid: An acid is a compound that reacts with water and increases the amount of hydronium ion present. In the chapter on acids and bases, we saw two more definitions of acids: a compound that donates a proton (a hydrogen ion, H+) to another compound is called a Brønsted-Lowry acid, and a Lewis acid is any species that can accept a pair of electrons. Explain why the introductory definition is a macroscopic definition, while the Brønsted-Lowry definition and the Lewis definition are microscopic definitions. Solution The introductory definition is a macroscopic definition in that it involves amounts of hydronium ion; amounts, such as concentrations, are macroscopic concepts. The other two definitions are microscopic because they are presented in terms of transfer of microscopic entities—a hydrogen ion or a pair of electrons.


Page 6 of 10

76. Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each: (a) –

2 3CO OH HCO (b) – –

3 4B(OH) OH B(OH) (c) –

2 3I I I (d) –

3 4AlCl Cl AlCl (use Al-Cl single bonds) (e) 2– 2

3 4O SO SO Solution

(a) ;

(b) ;

(c) ;

(d) ;

(e) 77. Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each: (a) –

2 3CS SH HCS (b) –

3 4BF F BF (c) –

2 3I SnI SnI (d) –

3 4Al(OH) OH Al(OH) (e) –

3 3F SO SFO Solution

(a) ;


Page 7 of 10

(b) ;

(c) ;

(d) ;

(e) 78. Using Lewis structures, write balanced equations for the following reactions: (a) 3HCl( ) PH ( ) g g (b) 3 3H O CH (c) 3CaO SO (d) –

4 2 5NH C H O Solution

(a) ; (b) 3 3 4 2H O + CH CH + H O

;


Page 8 of 10

(c) 3CaO + SO CaSO4

; (d) +

4 2 5 2 5 3NH + C H O C H OH + NH

79. Calculate 2

4[HgCl ] in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a 0.100-M HgCl2 solution. Solution The process is governed by the Kf of 2

4[HgCl ] . 2 – 2

4Hg ( ) 4Cl ( ) HgCl ( )aq aq aq 2

164f 2 4

[HgCl ] = = 1.1 10[Hg ][Cl ]

K

Because of the very large value of Kf, almost all the Hg2+ will form 24[HgCl ] . Let all the Hg2+

form 24[HgCl ] . In addition, let x be the change of 2

4[HgCl ] that dissociates:

16

f 4(0.070 ) = = 1.1 10

(0.030 )(4 )xK

x x

0.070 – x = 256x4(1.2 1015)(0.030 + x) Assume that x is small compared with 0.070 M and 0.030 M. x4 = 8.29 10–19 x = 3.02 10–5 The assumption was valid. Therefore:

24[HgCl ] = 0.070 − 3.02 10–5 = 0.070 M

80. In a titration of cyanide ion, 28.72 mL of 0.0100 M AgNO3 is added before precipitation begins. [The reaction of Ag+ with CN– goes to completion, producing the 2Ag(CN) complex.] Precipitation of solid AgCN takes place when excess Ag+ is added to the solution, above the amount needed to complete the formation of 2Ag(CN) . How many grams of NaCN were in the original sample? Solution


Page 9 of 10

The equilibrium is: + 21

2 fAg ( ) + 2CN ( ) Ag(CN) ( ) 1 10aq aq aq K � �� The number of moles of AgNO3 added is: 0.02872 L 0.0100 mol/L = 2.87 10–4 mol This compound reacts with CN– to form 2Ag(CN) , so there are 2.87 10–4 mol 2Ag(CN) . This amount requires 2 2.87 10–4 mol, or 5.74 10–4 mol, of CN–. The titration is stopped just as precipitation of AgCN begins:

+2AgCN ( ) + Ag ( ) 2AgCN( )aq aq s � ��

so only the first equilibrium is applicable. The value of Kf is very large. mol CN– < [ 2Ag(CN) ] mol NaCN = 2 mol [ 2Ag(CN) ] = 5.74 10–4 mol

4 49.007 gmass (NaCN) = 5.74 × 10 mol × = 0.0281 g1 mol

81. What are the concentrations of Ag+, CN–, and 2Ag(CN) in a saturated solution of AgCN? Solution Two equilibria are involved:

212 f

16sp

Ag 2CN [Ag(CN) ] 1 10

AgCN Ag CN 1.2 10

K

K

� ��

� ��

Solve the Ksp expression for [CN–] and substitute it into the first equation for Kf: Ksp = [Ag+][CN–] = 1.2 10–16

–161.2 10[CN ] = [Ag ]

21 212 232 2

2

21 16 112

[Ag(CN) ] [Ag(CN) ] 1 10 1 10[Ag ][1.44 10 ][Ag ][CN ]

[Ag ][Ag(CN) ][Ag ] 1 10 1.2 10 1.44 10

A relationship must be found to reduce the number of unknowns. The total amount of Ag+ in both Ag+ and 2Ag(CN) comes from AgCN only. Therefore:

22AgCN( ) Ag Ag(CN)s and

11 1/2 62[Ag ] [Ag(CN) ] (1.44 10 ) 3.8 10 M

From the Ksp, –16

11–6

1.2 10[CN ] = = 3.2 10 3.8 10

M

82. In dilute aqueous solution HF acts as a weak acid. However, pure liquid HF (boiling point = 19.5 C) is a strong acid. In liquid HF, HNO3 acts like a base and accepts protons. The acidity of liquid HF can be increased by adding one of several inorganic fluorides that are Lewis acids and accept F– ion (for example, BF3 or SbF5). Write balanced chemical equations for the reaction of pure HNO3 with pure HF and of pure HF with BF3.


Page 10 of 10

Solution –

3 2 3HNO ( ) HF( ) H NO Fl l ; 3 4HF( ) BF ( ) H BFl g 83. The simplest amino acid is glycine, H2NCH2CO2H. The common feature of amino acids is that they contain the functional groups: an amine group, –NH2, and a carboxylic acid group, –CO2H. An amino acid can function as either an acid or a base. For glycine, the acid strength of the carboxyl group is about the same as that of acetic acid, CH3CO2H, and the base strength of the amino group is slightly greater than that of ammonia, NH3. (a) Write the Lewis structures of the ions that form when glycine is dissolved in 1 M HCl and in1M KOH. (b) Write the Lewis structure of glycine when this amino acid is dissolved in water. (Hint: Consider the relative base strengths of the –NH2 and 2CO groups.) Solution

(a) ;

(b) 84. Boric acid, H3BO3, is not a Brønsted-Lowry acid but a Lewis acid. (a) Write an equation for its reaction with water. (b) Predict the shape of the anion thus formed. (c) What is the hybridization on the boron consistent with the shape you have predicted? Solution (a) 3 3 2 4 4H BO H O H BO H ; (b) First, form a symmetrical structure with the unique atom, B, as the central atom. Then include the 32e–to form the Lewis structure:

Because there are four bonds and no lone pair (unshared pair) on B, the electronic and molecular shapes are the same—both tetrahedral. (c) The tetrahedral structure is consistent with sp3 hybridization. This resource file is copyright 2019, Rice University. All Rights Reserved.

OpenStax Chemistry 2e 15.3: Coupled Equilibria

Page 1 of 9


15.3: Coupled Equilibria 85. A saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium? Solution The system described involves two reciprocal processes, dissolution and precipitation, occurring at equal rates. Because the equilibrium involves a solution and a solid, a difference in phases exists and the system is heterogeneous. 86. Calculate the equilibrium concentration of Ni2+ in a 1.0-M solution [Ni(NH3)6](NO3)2. Solution

2 2 83 3 6 fNi ( ) 6NH ( ) [Ni(NH ) ] ( ) 2.0 10aq aq aq K � ��

Let x be the change in concentration as Ni2+ dissociates. Because the initial Ni2+ concentration is 0, the concentration at any times is x:

28 3 6

2 6 63

[Ni(NH ) ] (1.0 )2.0 10 [Ni ][NH ] (6 )

xx x

2.0 108(46656x7) = 1.0 – x 9.33 1012(x2) = 1.0 – x Since x is small in comparison with 1.0, drop x: 9.33 1012(x7) = 1.0 x7 = 1.07 10–13 x = 0.014 M 87. Calculate the equilibrium concentration of Zn2+ in a 0.30-M solution of 2

4Zn(CN) . Solution The complex dissociates: 2– 2 –

4Zn(CN ) ( ) Zn ( ) 4CN ( )aq aq aq Let x be the change in concentration of Zn2+ that dissociates:

2

194f2 4

[Zn(CN) ] = = 2.1 10[Zn ][CN ]

K

194 5

0.30 0.30 = 2.1 10( )(4 ) 256

xx x x

≈ x = 3.5 10–5M = [Zn2+] 88. Calculate the equilibrium concentration of Cu2+ in a solution initially with 0.050 M Cu2+ and 1.00 M NH3. Solution Assume that all Cu2+ forms the complex whose concentration is 0.050 M and the remaining NH3 has a concentration of 1.00 M – 4(0.050 M) = 0.80 M. The complex dissociates:


Page 2 of 9

2 23 4 3[Cu(NH ) ] [Cu ] 4[NH ] � ��

Let x be the change in concentration of Cu2+ that dissociates:

2

133 42 4 4

3

[Cu(NH ) ] 0.050 1.7 10 [Cu ][NH ] (4 0.80)

xx x

Assume that 4x is small when compared with 0.80 and that x is small when compared with 0.050: (0.80)4 1.7 1013 x = 0.050 x = 7.2 10–15M 89. Calculate the equilibrium concentration of Zn2+ in a solution initially with 0.150 M Zn2+ and 2.50 M CN–. Solution Assume that all Zn2+ forms the complex whose concentration is 0.150 M and [CN–] is 2.50 M – 4(0.150 M) = 1.90 M. Let x be the change in concentration of Zn2+ that dissociates:

2

1942 4 4

[Zn(CN) ] 0.150 = 2.1 10 = [Zn ][CN ] (4 1.90)

xx x

Assume that x is small when compared with 0.150 and that 4x is small when compared with 1.90:

19f 4

0.150 = = 2.1 10(1.90)

Kx

x = [Zn2+] = 5.5 10–22M 90. Calculate the Fe3+ equilibrium concentration when 0.0888 mole of K3[Fe(CN)6] is added to a solution with 0.0.00010 M CN–. Solution Set up a table listing initial and equilibrium concentrations for the reaction:

3 3 436 fFe 6CN (Fe(CN) ] 2 10K � ��

Let x be the concentration of Fe3+ that dissociates when 0.0888 mol dissolves in 1.00 L of 0.00010 M CN–. Assume no volume change upon dissolution:


Page 3 of 9

3436

3 6 6

[Fe(CN) ] 0.0888 2 10[Fe ][CN ] (0.000010 6 )

xx x

Assume that x is small when compared with the terms from which it is subtracted: 0.0888 = (0.00010)6(x)(1 1043)

2219

0.0888 = = 4.4 10 2 10

x M

91. Calculate the Co2+ equilibrium concentration when 0.100 mole of [Co(NH3)6](NO3)2 is added to a solution with 0.025 M NH3. Assume the volume is 1.00 L. Solution Assume that 1.00 L of solution is present and that addition of 0.100 mol will not change the volume.

2 23 3 6Co + 6NH [Co(NH ) ] � ��

Let Δ = concentration of 23 6[Co(NH ) ] that dissolves.

Kf = 2

53 6f 2 6

3

[Co(NH ) ] (0.100 )= = 1.3 × 10 = [Co ][NH ] (0.025 + 6 )

K

Assume that Δ is small compared with 0.025 and 0.100. Δ = 3.1 10–5

Check: 46 1.86 × 10 100% = 100% = 0.7%

0.025 0.025

The assumption is justified and the value is [Co2+] = 4.8 10–5 M 92. Calculate the molar solubility of Sn(OH)2 in a buffer solution containing equal concentrations ofNH3 and 4NH . Solution

54b

3

[NH ][OH ] 1.8 10[NH ]

K

because 4[NH ] = [NH3],

[OH–] = Kb = 1.8 10–5 M [Sn2+](1.8 10–5)2 = 3 10–27 [Sn2+] = 9 10–18 M (molar solubility) 93. Calculate the molar solubility of Al(OH)3 in a buffer solution with 0.100 M NH3 and 0.400 M

4NH . Solution

54b

3

[NH ][OH ] (0.400)[OH ] = = = 1.8 10[NH ] (0.100)

K

55(0.100)(1.8 10 )[OH ] = = 4.5 10

0.0400

Ksp = [Al3+][OH–]3 = [Al3+](4.5 10–5)3 = 2 10–32 [Al3+] = 2 10–19 (molar solubility) 94. What is the molar solubility of CaF2 in a 0.100-M solution of HF? Ka for HF = 6.4 10–4. Solution


Page 4 of 9

Find the concentration of F– present from the given Ka: 4

a[H ][F ] ( )( ) 6.4 10

[HF] 0.100 x xK

x

where x is the concentration of F– formed. Assume that x is small when compared with 0.100 and can be neglected. x2 = 0.100(6.4 10–4) = 6.4 10–5 x = 8.0 10–3 = [F–] Check the assumption:

38.0 10 100% 8.0%0.100

This percentage error is too large, so a quadratic equation must be solved: x2 + 6.4 10–4x – 6.4 10–5 = 0 Of the two roots, only one is positive: x = 0.0077 Ksp = [Ca2+][F–]2 = [Ca2+](0.0077)2 = 4.0 10–11 [Ca2+] = 6.7 10–7 M (molar solubility) 95. What is the molar solubility of BaSO4 in a 0.250-M solution of NaHSO4? Ka for 4HSO = 1.2 10–2. Solution Find the amount of 2

4SO present from Ka for the equilibrium: 2

4 2 3 4HSO H O H O SO Let x be 2

4[SO ] : 2 2

23 4a

4

[H O ][SO ] = = = 1.2 10[HSO ] 0.250

xKx

Because Ka is too large to disregard x in the expression 0.250 – x, we must solve the quadratic equation: x2 + 1.2 10–2x – 0.250(1.2 10–2) = 0

2 2 2 3 21.2 10 (1.2 10 ) 4(3.0 10 ) 1.2 10 0.11 = = = 0.049 2 2

x M

Ksp = [Ba2+] 24[SO ] = [Ba2+](0.049) = 2.3 10–8

[Ba2+] = 4.7 10–7 (molar solubility) 96. What is the molar solubility of Tl(OH)3 in a 0.10-M solution of NH3? Solution [OH–] is governed by the equilibrium:

3 2 4

54b

3

NH H O NH OH

[NH ][OH ] ( )( ) 1.8 10[NH ] 0.10

x xKx

� ��

Assume that x is small when compared with 0.10: x2 = 0.10 1.8 10–5 = 1.8 10–6 x = 1.34 10–3 Ksp = [Tl3+][OH–] = [Tl3+][1.34 10–3]3 = 6.3 10–46


Page 5 of 9

[Tl3+] = 2.6 10–37 (molar solubility) 97. What is the molar solubility of Pb(OH)2 in a 0.138-M solution of CH3NH2? Solution

+ –3 2 2 3 3CH NH H O CH NH OH � ��

43 3b

3 2

[CH NH ][OH ] ( )( ) = = = 4.4 10[CH NH ] 0.138

x xKx

Solve the quadratic equation using the quadratic formula: x2 + 4.4 10–4x – 0.138(4.4 10–4) = 0

4 4 2 5 4 4

3

4.4 10 (4.4 10 ) 4(6.07 10 ) 4.4 10 2.43 10 = = 2 2

0.0152= = 7.6 10 2

x

M

Ksp = [Pb2+][OH–]2 = [Pb2+](7.6 10–3)2 = 1.2 10–15 [Pb2+] = 2.1 10–11 (molar solubility) 98. A solution of 0.075 M CoBr2 is saturated with H2S ([H2S] = 0.10 M). What is the minimum pH at which CoS begins to precipitate?

2 2– –26spCoS( ) Co ( ) S ( ) 3 10s aq aq K � ��


Solution Two equilibria are in competition for the ions and must be considered simultaneously. Precipitation of CoS will occur when the concentration of S2– in conjunction with 0.075 M Co2+ exceeds Ksp of CoS. The [S2–] must come from the ionization of H2S as defined by the equilibrium:

22 2 3

2273

1 2 22

H S( ) 2H O( ) 2H O ( ) S ( )

[H O ][S ] (H S) 8.9 10[H S]

aq l aq aq

K K

� ��

As a saturated solution of H2S is 0.10 M, this expression becomes: +

3[H O ] [S2–] = 8.9 10–28 From the equilibrium of CoS, the minimum concentration of S2– required to cause precipitation is calculated as:

2 2CoS( ) Co ( ) S ( )s aq aq � �� Ksp = [Co2+][S2–] = 3.0 10–26

262 253 10[S ] 4.0 10

0.075

This amount of S2– will exist in solution at a pH defined by the H2S equilibrium: + 2 25 28

3[H O ] (4.0 10 ) = 8.9 10 + 2 3

3[H O ] = 2.225 10 + 2

3[H O ] = 4.72 10 pH = –log 4.72 10–2 = 1.33


Page 6 of 9

99. A 0.125-M solution of Mn(NO3)2 is saturated with H2S ([H2S] = 0.10 M). At what pH does MnS begin to precipitate?

2 2– –13spMnS( ) Mn ( ) S ( ) 2.3 10s aq aq K � ��


Solution Two equilibria are in competition for the ions and must be considered simultaneously. Precipitation of MnS will occur when the concentration of S2– in conjunction with 0.125 M Mn2+ exceeds the Ksp of MnS. The [S2–] must come from the ionization of H2S as defined by the equilibrium:

2–2 2 3H S( ) 2H O( ) 2H O ( ) S ( )aq l aq aq

2 2273

1 2 22

[H O ] [S ] = (H S) = 8.9 10

[H S]K K

As a saturated solution of H2S is 0.10 M, this later expression becomes: + 2 2 28

3[H O ] [S ] = 8.9 10 From the equilibrium of MnS, the minimum concentration of S2– required to cause precipitation is calculated as:

2 2–MnS( ) Mn ( ) S ( )s aq aq Ksp = [Mn2+][S2–] = 2.3 10–13

132 122.3 10[S ] = = 1.84 10

0.125

This amount of S2– will exist in solution at a pH defined by the H2S equilibrium: + 2 12 28

3[H O ] (1.84 10 ) = 8.9 10 2 16

3[H O ] 4.84 10 + 8

3[H O ] = 2.20 10 M +

3pH = log[H O ] = 7.66 100. Both AgCl and AgI dissolve in NH3. (a) What mass of AgI dissolves in 1.0 L of 1.0 M NH3? (b) What mass of AgCl dissolves in 1.0 L of 1.0 M NH3? Solution (a) There are two coupled equilibrium processes occurring in these solutions: dissolution and complex ion formation. First, combine the two equilibrium equations to yield an equation for the overall reaction:

16spAgI( ) Ag ( ) I ( ) 1.5 10s aq aq K � ��

73 3 2 fAg ( ) 2NH ( ) Ag(NH ) ( ) 1.7 10aq aq aq K � ��

There are two coupled equilibrium processes occurring in these solutions: dissolution and complex ion formation. First, combine the two equilibrium equations to yield an equation for the overall reaction:

93 3 2 sp fAgI( ) + 2NH ( ) Ag(NH ) ( ) I ( ) 2.6 10s aq aq aq K K K � ��

Next, use the ICE approach to compute the concentration of the dissolved complex ion at equilibrium:


Page 7 of 9

K = [Ag(NH3)+(aq)][ I–(aq)] / [NH3]2 = (x)(x) / 1.0 – 2x Since K is very small, assume 2x << 1.0, then simplify and solve the equation for x: 2.6 × 10–9 = x2 / 1.0 x = [Ag(NH3)+(aq)] = [ I–(aq)] = 5.05 × 10–5 M The mass (g) of AgI that must dissolve to yield these ion concentrations is

5 2234.772 g5.0 10 mol 1.2 10 g AgI1 mol

.

(b) The same approach is followed as in part (a), but in this case the equilibrium constant for the overall reaction is not very small (K = 0.0027) and so the simplifying assumption that 2x << 1.0 is not valid. This requires solving the resulting quadratic equation: x2 + 0.0054x – 0.0027 = 0 x = [Ag(NH3)+(aq)] = [ I–(aq)] = 0.049 M The mass (g) of AgCl that must dissolve to yield these ion concentrations is (0.0049 mol/L)(143.321 g/mol) = 7.0 g AgCl 101. The following question is taken from a Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service. Solve the following problem:

2 –2MgF ( ) Mg ( ) 2F ( )s aq aq � ��

In a saturated solution of MgF2 at 18C, the concentration of Mg2+ is 1.21 10–3 M. The equilibrium is represented by the equation above. (a) Write the expression for the solubility-product constant, Ksp, and calculate its value at 18 C. (b) Calculate the equilibrium concentration of Mg2+ in 1.000 L of saturated MgF2 solution at 18 C to which 0.100 mol of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible. (c) Predict whether a precipitate of MgF2 will form when 100.0 mL of a 3.00 10–3-M solution of Mg(NO3)2 is mixed with 200.0 mL of a 2.00 10–3-M solution of NaF at 18 C. Show the calculations to support your prediction. (d) At 27 C the concentration of Mg2+ in a saturated solution of MgF2 is 1.17 10–3M. Is the dissolving of MgF2 in water an endothermic or an exothermic process? Give an explanation to support your conclusion. Solution (a) Ksp = [Mg2+][F–]2 = (1.21 10–3)(2 1.21 10–3)2 = 7.09 10–9; (b) Ksp = [Mg2+][F–]2 = [x][0.100 + 2x]2 = 7.09 10–9 Assume that 2x is small when compared with 0.100 M. 0.100x = 7.09 10–9 x = [MgF2] = 7.09 10–7 M The value 7.09 10–7 M is quite small when compared with 0.100 M, so the assumption is valid.


Page 8 of 9

(c) Determine the concentration of Mg2+ and F– that will be present in the final volume. Compare the value of the ion product [Mg2+][F–]2 with Ksp. If this value is larger than Ksp, precipitation will occur. 0.1000 L 3.00 10–3 M Mg(NO3)2 = 0.3000 L M Mg(NO3)2 M Mg(NO3)2 = 1.00 10–3 M 0.2000 L 2.00 10–3 M NaF = 0.3000 L M NaF M NaF = 1.33 10–3 M ion product = (1.00 10–3)(1.33 10–3)2 = 1.77 10–9 This value is smaller than Ksp, so no precipitation will occur. (d) MgF2 is less soluble at 27 °C than at 18 °C. Because added heat acts like an added reagent, when it appears on the product side, the Le Châtelier’s principle states that the equilibrium will shift to the reactants’ side to counter the stress. Consequently, less reagent will dissolve. This situation is found in our case. Therefore, the reaction is exothermic. 102. Which of the following compounds, when dissolved in a 0.01-M solution of HClO4, has a solubility greater than in pure water: CuCl, CaCO3, MnS, PbBr2, CaF2? Explain your answer. Solution CaCO3, MnS, CaF2; each is a salt of a weak acid and the +

3[H O ] from perchloric acid reduces the equilibrium concentration of the anion 103. Which of the following compounds, when dissolved in a 0.01-M solution of HClO4, has a solubility greater than in pure water: AgBr, BaF2, Ca3(PO4)3, ZnS, PbI2? Explain your answer. Solution BaF2, Ca3(PO4)2, ZnS; each is a salt of a weak acid, and the +

3[H O ] from perchloric acid reduces the equilibrium concentration of the anion, thereby increasing the concentration of the cations 104. What is the effect on the amount of solid Mg(OH)2 that dissolves and the concentrations of Mg2+ and OH– when each of the following are added to a mixture of solid Mg(OH)2 and water at equilibrium? (a) MgCl2 (b) KOH (c) HClO4 (d) NaNO3 (e) Mg(OH)2? Solution Effect on amount of solid Mg(OH)2, [Mg2+], [OH–]: (a) increase, increase, decrease; (b) increase, decrease, increase; (c) decrease, increase, decrease; (d) no effect predicted; (e) increase, no effect, no effect 105. What is the effect on the amount of CaHPO4 that dissolves and the concentrations of Ca2+ and 4HPO when each of the following are added to a mixture of solid CaHPO4 and water at equilibrium? (a) CaCl2 (b) HCl (c) KClO4 (d) NaOH (e) CaHPO4 Solution


Page 9 of 9

Effect on amount of solid CaHPO4, [Ca2+], [OH–]: (a) increase, increase, decrease; (b) decrease, increase, decrease; (c) no effect, no effect, no effect; (d) decrease, increase, decrease; (e) increase, no effect, no effect 106. Identify all chemical species present in an aqueous solution of Ca3(PO4)2 and list these species in decreasing order of their concentrations. (Hint: Remember that the 3

4PO ion is a weak base.) Solution [H2O] > [Ca2+] > 3

4[PO ] > 2–4HPO = [OH–] > 4[HPO ]

This resource file is copyright 2019, Rice University. All Rights Reserved.

openstax chemistry 2e

Documents