19-1 chem 1b: general chemistry chapter 19: ionic equilibria in aqueous systems instructor: dr....

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CHEM 1B:GENERAL CHEMISTRY

Chapter 19: Ionic Equilibria in Aqueous Systems

Instructor: Dr. Orlando E. RaolaSanta Rosa Junior College

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Chapter 19

Ionic Equilibria in Aqueous Systems

19-3

Ionic Equilibria in Aqueous Systems

19.1 Equilibria of Acid-Base Buffers

19.2 Acid-Base Titration Curves

19.3 Equilibria of Slightly Soluble Ionic Compounds

19.4 Equilibria Involving Complex Ions

19-4

Acid-Base Buffers

An acid-base buffer is a solution that lessens the impact of pH from the addition of acid or base.

An acid-base buffer usually consists of a conjugate acid-base pair where both species are present in appreciable quantities in solution.

An acid-base buffer is therefore a solution of a weak acid and its conjugate base, or a weak base and its conjugate acid.

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Figure 19.1 The effect of adding acid or base to an unbuffered solution.

A 100-mL sample of dilute HCl is adjusted to pH 5.00.

The addition of 1 mL of strong acid (left) or strong base (right) changes the pH by several units.

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Figure 19.2 The effect of adding acid or base to a buffered solution.

A 100-mL sample of an acetate buffer is adjusted to pH 5.00.

The addition of 1 mL of strong acid (left) or strong base (right) changes the pH very little.

The acetate buffer is made by mixing 1 mol/L CH3COOH ( a weak acid) with 1 mol/L CH3COONa (which provides the conjugate base, CH3COO-).

19-7

Buffers and the Common-ion Effect

A buffer works through the common-ion effect.

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

acetic acid acetate ion

Acetic acid in water dissociates slightly to produce some acetate ion:

If NaCH3COO is added, it provides a source of CH3COO- ion, and the equilibrium shifts to the left. CH3COO- is common to both solutions.

The addition of CH3COO- reduces the % dissociation of the acid.

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Table 19.1 The Effect of Added Acetate Ion on the Dissociation of Acetic Acid

[CH3COOH]init [CH3COO-]added % Dissociation* [H3O+] pH

0.10 0.00 1.3 1.3x10-3 2.89

0.10 0.050 0.036 3.6x10-5 4.44

0.10 0.10 0.018 1.8x10-5 4.74

0.10 0.15 0.012 1.2x1015 4.92

* % Dissociation = [CH3COOH]dissoc

[CH3COOH]init

x 100

19-9

How a Buffer Works

The buffer components (HA and A-) are able to consume small amounts of added OH- or H3O+ by a shift in equilibrium position.


Added H3O+ reacts with CH3COO-, causing a shift to the left.

Added OH- reacts with CH3COOH, causing a shift to the right.

The shift in equilibrium position absorbs the change in [H3O+] or [OH-], and the pH changes only slightly.

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Figure 19.3 How a buffer works.

Buffer has equal concentrations of A- and HA.

H3O+

Buffer has more HA after addition of H3O+.

H2O + CH3COOH ← H3O+ + CH3COO-

OH-

Buffer has more A- after addition of OH-.

CH3COOH + OH- → CH3COO- + H2O

19-11

Relative Concentrations of Buffer Components


Ka =[CH3COO-][H3O+]

[CH3COOH]

If the ratio increases, [H3O+] decreases.[HA]

[A-]

Since Ka is constant, the [H3O+] of the solution depends on the ratio of buffer component concentrations.

[H3O+] = Ka x[CH3COOH]

[CH3COO-]

If the ratio decreases, [H3O+] increases.[HA]

[A-]

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Sample Problem 19.1 Calculating the Effect of Added H3O+ or OH- on Buffer pH

Calculate the pH:

PLAN: We can calculate [CH3COOH]init and [CH3COO-]init from the given information. From this we can find the starting pH. For (b) and (c) we assume that the added OH- or H3O+ reacts completely with the buffer components. We write a balanced equation in each case, set up a reaction table, and calculate the new [H3O+].

PROBLEM:

(a) Of a buffer solution consisting of 0.50 mol/L CH3COOH and 0.50 mol/L CH3COONa

(b) After adding 0.020 mol of solid NaOH to 1.0 L of the buffer solution

(c) After adding 0.020 mol of HCl to 1.0 L of the buffer solution in (a).

Ka of CH3COOH = 1.8 x 10-5. (Assume the additions cause a negligible change in volume.)

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Sample Problem 19.1

SOLUTION: (a)

Initial 0.50 -

0.50 0

Change -−x +x +x

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) Concentration (mol/L)

Equilibrium - 0.50 - x x0.50 + x

1.8x10-5 mol/L

0.50 mol/Lx 100 = 3.6x10-3% (< 5%; assumption is justified.)

Since Ka is small, x is small, so we assume

[CH3COOH] = 0.50 – x ≈ 0.50 mol/L and [CH3COO-] = 0.50 + x ≈ 0.50 mol/L

x = [H3O+] = Ka x[CH3COOH]

[CH3COO-]≈ 1.8x10-5 x 0.50

0.50= 1.8x10-5 mol/L

Checking the assumption: pH = -log(1.8x10-5) = 4.74

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Sample Problem 19.1

Initial 0.50 0.020 0.50 -

Change -0.020-0.020 +0.020 -

CH3COOH(aq) + OH-(aq) → CH3COO-(aq) + H2O(l) Concentration (mol/L)

(b) [OH−]added = 0.020 mol

1.0 L soln= 0.020 mol/L OH−

Equilibrium 00.48 -0.52

Setting up a reaction table for the stoichiometry:

Initial 0.48 -

0.52 0

Change -−x +x +x



Setting up a reaction table for the acid dissociation, using new initial [ ]:

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Sample Problem 19.1




[CH3COO-]

pH = -log(1.7x10-5) = 4.77

≈ 1.8x10-5 x 0.48

0.52= 1.7x10-5 mol/L

19-16

Sample Problem 19.1

Initial 0.50 0.020 0.50 -

Change -0.020-0.020 +0.020 -

CH3COO-(aq) + H3O+(aq) → CH3COOH(aq) + H2O(l) Concentration (mol/L)

(c) [H3O+]added = 0.020 mol

1.0 L soln= 0.020 mol/L H3O+

Equilibrium 00.48 -0.52

Setting up a reaction table for the stoichiometry:

Initial 0.52 -

0.48 0

Change -−x +x +x



Setting up a reaction table for the acid dissociation, using new initial [ ]:

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Sample Problem 19.1




[CH3COO-]

pH = -log(2.0x10-5) = 4.70

≈ 1.8x10-5 x 0.52

0.48= 2.0x10-5 mol/L

19-18

The Henderson-Hasselbalch Equation

HA(aq) + H2O(l) A-(aq) + H3O+(aq)

Ka =[H3O+][A-]

[HA][H3O+] = Ka x

[HA]

[A-]

-log[H3O+] = -logKa – log [HA]

[A-]

pH = pKa + log[base]

[acid]

19-19

Buffer Capacity

The buffer capacity is a measure of the “strength” of the buffer, its ability to maintain the pH following addition of strong acid or base.

The greater the concentrations of the buffer components, the greater its capacity to resist pH changes.

The closer the component concentrations are to each other, the greater the buffer capacity.

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Figure 19.4 The relation between buffer capacity and pH change.

When strong base is added, the pH increases least for the most concentrated buffer.

This graph shows the final pH values for four different buffer solutions after the addition of strong base.

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Buffer Range

The buffer range is the pH range over which the buffer is effective.

Buffer range is related to the ratio of buffer component concentrations.

[HA]

[A-]The closer is to 1, the more effective the buffer.

If one component is more than 10 times the other, buffering action is poor. Since log10 = 1, buffers have a usable range within ± 1 pH unit of the pKa of the acid component.

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Sample Problem 19.2 Using Molecular Scenes to Examine Buffers

PROBLEM: The molecular scenes below represent samples of four HA/A- buffers. (HA is blue and green, A- is green, and other ions and water are not shown.)

(a) Which buffer has the highest pH?(b) Which buffer has the greatest capacity?(c) Should we add a small amount of concentrated strong acid or

strong base to convert sample 1 to sample 2 (assuming no volume changes)?

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Sample Problem 19.2

PLAN: Since the volumes of the solutions are equal, the scenes represent molarities as well as numbers. We count the particles of each species present in each scene and calculate the ratio of the buffer components.

SOLUTION:

[A-]/[HA] ratios: sample 1, 3/3 = 1; sample 2, 2/4 = 0.5; sample 3, 4/4 = 1; and sample 4, 4/2 = 2.

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Sample Problem 19.2

(c) Sample 2 has a lower [A-]/[HA] ratio than sample 1, so we need to increase the [A-] and decrease the [HA]. This is achieved by adding strong acid to sample 1.

(a) As the pH rises, more HA will be converted to A-. The scene with the highest [A-]/[HA] ratio is at the highest pH. Sample 4 has the highest pH because it has the highest ratio.

(b) The buffer with the greatest capacity is the one with the [A-]/[HA] closest to 1. Sample 3 has the greatest buffer capacity.

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Preparing a Buffer

• Choose the conjugate acid-base pair.– The pKa of the weak acid component should be close to the

desired pH.

• Calculate the ratio of buffer component concentrations.

• Determine the buffer concentration, and calculate the required volume of stock solutions and/or masses of components.

• Mix the solution and correct the pH.


[acid]

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Sample Problem 19.3 Preparing a Buffer

SOLUTION:

PROBLEM: An environmental chemist needs a carbonate buffer of pH 10.00 to study the effects of the acid rain on limsetone-rich soils. How many grams of Na2CO3 must she add to 1.5 L of freshly prepared 0.20 mol/L NaHCO3 to make the buffer? Ka of HCO3

- is 4.7x10-11.

PLAN: The conjugate pair is HCO3- (acid) and CO3

2- (base), and we know both the buffer volume and the concentration of HCO3

-. We can calculate the ratio of components that gives a pH of 10.00, and hence the mass of Na2CO3 that must be added to make 1.5 L of solution.

[H3O+] = 10-pH = 10-10.00 = 1.0x10-10 mol/L

HCO3-(aq) + H2O(l) H3O+(aq) + CO3

2-(aq) Ka =[CO3

2-][H3O+]

[HCO3-]

19-27

Sample Problem 19.3 Preparing a Buffer

[CO32-] =

Ka[HCO3-]

[H3O+]=

(4.7x10-11)(0.20)

1.0x10-10= 0.094 mol/L

Amount (mol) of CO32- needed = 1.5 L soln x 0.094 mol CO3

2-

1 L soln

= 0.14 mol CO32-

105.99 g Na2CO3

1 mol Na2CO3

0.14 mol Na2CO3 x = 15 g Na2CO3

The chemist should dissolve 15 g Na2CO3 in about 1.3 L of 0.20 mol/L NaHCO3

and add more 0.20 mol/L NaHCO3 to make 1.5 L. Using a pH meter, she can then adjust the pH to 10.00 by dropwise addition of concentrated strong acid or base.

19-28

Acid-Base Indicators

An acid-base indicator is a weak organic acid (HIn) whose color differs from that of its conjugate base (In-).

The color of an indicator changes over a specific, narrow pH range, a range of about 2 pH units.

The ratio [HIn]/[In-] is governed by the [H3O+] of the solution. Indicators can therefore be used to monitor the pH change during an acid-base reaction.

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pH

Figure 19.5 Colors and approximate pH range of some common acid-base indicators.

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Figure 19.6 The color change of the indicator bromthymol blue.

pH < 6.0 pH > 7.5pH = 6.0-7.5

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Acid-Base Titrations

In an acid-base titration, the concentration of an acid (or a base) is determined by neutralizing the acid (or base) with a solution of base (or acid) of known concentration.

The equivalence point of the reaction occurs when the number of moles of OH- added equals the number of moles of H3O+ originally present, or vice versa.

The end point occurs when the indicator changes color.- The indicator should be selected so that its color change occurs at a pH close to that of the equivalence point.

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Figure 19.7 Curve for a strong acid–strong base titration.

The initial pH is low.

The pH rises very rapidly at the equivalence point, which occurs at pH = 7.00.

The pH increases gradually when excess base has been added.

19-33

Calculating the pH during a strong acid–strong base titration

Initial pH

[H3O+] = [HA]init

pH = -log[H3O+]

pH before equivalence point

initial mol H3O+ = Vacid x Macid

mol OH- added = Vbase x Mbase

mol H3O+remaining = (mol H3O+

init) – (mol OH-added)

[H3O+] = pH = -log[H3O+]mol H3O+

remaining

Vacid + Vbase

19-34

pH at the equivalence point

pH = 7.00 for a strong acid-strong base titration.

Calculating the pH during a strong acid–strong base titration

pH beyond the equivalence point

initial mol H3O+ = Vacid x Macid

mol OH- added = Vbase x Mbase

mol OH-excess = (mol OH-

added) – (mol H3O+init)

[OH-] =

pOH = -log[OH-] and pH = 14.00 - pOH

mol OH-excess

Vacid + Vbase

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Example:

40.00 mL of 0.1000 mol/L HCl is titrated with 0.1000 mol/L NaOH.

The initial pH is simply the pH of the HCl solution:

[H3O+] = [HCl]init = 0.1000 mol/L and pH = -log(0.1000) = 1.00

To calculate the pH after 20.00 mL of NaOH solution has been added:

OH- added = 0.02000 L NaOH x 0.1000 mol

1 L= 2.000x10-3 mol OH-

Initial mol of H3O+ = 0.04000 L HCl x 0.1000 mol

1 L= 4.000x10-3 mol H3O+

H3O+ remaining = 4.000x10-3 – 2.000x10-3 = 2.000x10-3 mol H3O+

The OH- ions react with an equal amount of H3O+ ions, so

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[H3O+] = 2.000x10-3 mol

0.04000 L + 0.02000 L= 0.03333 mol/L

pH = -log(0.03333) = 1.48

To calculate the pH after 50.00 mL of NaOH solution has been added:

The equivalence point occurs when mol of OH- added = initial mol of HCl, so when 40.00 mL of NaOH has been added.

OH- added = 0.05000 L NaOH x 0.1000 mol

1 L= 5.000x10-3 mol OH-

OH- in excess = 5.000x10-3 – 4.000x10-3 = 1.000x10-3 mol OH-

[OH-] = 1.000x10-3 mol

0.04000 L + 0.05000 L= 0.01111 mol/L

pOH = -log(0.01111) = 1.95 pH = 14.00 – 1.95 = 12.05

19-37

Figure 19.8 Curve for a weak acid–strong base titration.

The initial pH is higher than for the strong acid solution.

The curve rises gradually in the buffer region. The weak acid and its conjugate base are both present in solution.

The pH increases slowly beyond the equivalence point.

The pH at the equivalent point is > 7.00 due to the reaction of the conjugate base with H2O.

19-38

Calculating the pH during a weak acid–strong base titration

Initial pH

[H3O+] =

pH = -log[H3O+]

[H3O+][A-]

[HA]Ka = Ka x [HA]init)

pH before equivalence point

[H3O+] = Ka x or


[acid]

[HA]

[A-]

19-39

Calculating the pH during a weak acid–strong base titration

A-(aq) + H2O(l) HA(aq) + OH-(aq)

pH at the equivalence point

[OH- ] =

where [A-] = andKw

Ka

Kb =mol HAinit

Vacid + Vbase

Kb x [A-]

[H3O+] ≈ and pH = -log[H3O+]Kw

Kb x [A-])

pH beyond the equivalence point

[OH-] =

pH = -log[H3O+]

mol OH-excess

Vacid + Vbase

[H3O+] = Kw

[OH-]

19-40

Sample Problem 19.4 Finding the pH During a Weak Acid–Strong Base Titration

PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000 mol/L propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of 0.1000 mol/L NaOH:

PLAN: The initial pH must be calculated using the Ka value for the weak acid. We then calculate the number of moles of HPr present initially and the number of moles of OH- added. Once we know the volume of base required to reach the equivalence point we can calculate the pH based on the species present in solution.

(a) 0.00 mL (b) 30.00 mL (c) 40.00 mL (d) 50.00 mL

SOLUTION:

(a) [H3O+] = Ka x [HPr]init) = (1.3x10-5)(0.100) = 1.1x10-3 mol/L

pH = -log(1.1x10-3) = 2.96

19-41

Sample Problem 19.4

(b) 30.00 mL of 0.1000 mol/L NaOH has been added.

Initial amount of HPr = 0.04000 L x 0.1000 mol/L = 4.000x10-3 mol HPr

Amount of NaOH added = 0.03000 L x 0.1000 mol/L = 3.000x10-3 mol OH-

Each mol of OH- reacts to form 1 mol of Pr-, so

HPr(aq) + OH-(aq) → Pr-(aq) + H2O(l)Concentration (mol/L)Initial 0.004000 0.003000

0 -

Change −0.003000 -0.003000

+0.003000

-

Equilibrium 0.001000 0 -0.003000

[H3O+] = Ka x [HPr]

[Pr-]

pH = -log(4.3x10-6) = 5.37

0.001000

0.003000= (1.3x10-5) x = 4.3x10-6 mol/L

19-42

Sample Problem 19.4

(c) 40.00 mL of 0.1000 mol/L NaOH has been added.

This is the equivalence point because mol of OH- added = 0.004000 = mol of HAinit. All the OH- added has reacted with HA to form 0.004000 mol of Pr-.

0.004000 mol

0.04000 L + 0.04000 L[Pr-] = = 0.05000 mol/L

Pr- is a weak base, so we calculate Kb =Kw

Ka

=1.0x10-14

1.3x10-5= 7.7x10-10

[H3O+] ≈Kw

Kb x [A-])=

(7.7x10-10)(0.05000)1.0x10-14

= 1.6x10-9 mol/L

pH = -log(1.6x10-9) = 8.80

19-43

Sample Problem 19.4

(d) 50.00 mL of 0.1000 mol/L NaOH has been added.

Amount of OH- added = 0.05000 L x 0.1000 mol/L = 0.005000 mol

Excess OH- = OH-added – HAinit = 0.005000 – 0.004000 = 0.001000 mol

[OH-] = mol OH-

excess

total volume=

0.001000 mol

0.09000 L= 0.01111 mol/L

[H3O+] = Kw

[OH-]=

1x10-14

0.01111= 9.0x10-13 mol/L

pH = -log(9.0x10-13) = 12.05

19-44

Figure 19.9 Curve for a weak base–strong acid titration.

The pH decreases gradually in the buffer region. The weak base and its conjugate acid are both present in solution.

The pH at the equivalence point is < 7.00 due to the reaction of the conjugate acid with H2O.

19-45

Figure 19.10 Curve for the titration of a weak polyprotic acid.

Titration of 40.00 mL of 0.1000 mol/L H2SO3 with 0.1000 mol/L

NaOH

pKa2 = 7.19

pKa1 = 1.85

19-46

Amino Acids as Polyprotic Acids

An amino acid contains a weak base (-NH2) and a weak acid (-COOH) in the same molecule.

Both groups are protonated at low pH and the amino acid behaves like a polyprotic acid.

H2N C COOH

H

R

The general formula for an amino acid

19-47

Figure 19.11 Abnormal shape of red blood cells in sickle cell anemia.

In sickle cell anemia, the hemoglobin has two amino acids with neutral R groups instead of charged groups. The abnormal hemoglobin causes the red blood cells to have a sickle shape, as seen here.

Several amino acids have charged R groups in addition to the NH2 and COOH group. These are essential to the normal structure of many proteins.

19-48

Equilibria of Slightly Soluble Ionic Compounds

Any “insoluble” ionic compound is actually slightly soluble in aqueous solution.We assume that the very small amount of such a compound that dissolves will dissociate completely.

For a slightly soluble ionic compound in water, equilibrium exists between solid solute and aqueous ions.

PbF2(s) Pb2+(aq) + 2F-(aq)

Qc =[Pb2+][F-]2

[PbF2]Qsp = Qc[PbF2] = [Pb2+][F-]2

19-49

Qsp and Ksp

Qsp is called the ion-product expression for a slightly soluble ionic compound.

For any slightly soluble compound MpXq, which consists of ions Mn+ and Xz-,

Qsp = [Mn+]p[Xz-]q

When the solution is saturated, the system is at equilibrium, and Qsp = Ksp, the solubility product constant.

The Ksp value of a salt indicates how far the dissolution proceeds at equilibrium (saturation).

19-50

Metal Sulfides

Metal sulfides behave differently from most other slightly soluble ionic compounds, since the S2- ion is strongly basic.

We can think of the dissolution of a metal sulfide as a two-step process:

MnS(s) Mn2+(aq) + S2-(aq)

S2-(aq) + H2O(l) → HS-(aq) + OH-(aq)

MnS(s) + H2O(l) Mn2+(aq) + HS-(aq) + OH-(aq)

Ksp = [Mn2+][HS-][OH-]

19-51

Sample Problem 19.5 Writing Ion-Product Expressions

SOLUTION:

PROBLEM: Write the ion-product expression at equilibrium for each compound:(a) magnesium carbonate (b) iron(II) hydroxide(c) calcium phosphate (d) silver sulfide

PLAN: We write an equation for a saturated solution of each compound, and then write the ion-product expression at equilibrium, Ksp. Note the sulfide in part (d).

(a) MgCO3(s) Mg2+(aq) + CO32-(aq) Ksp = [Mg2+][CO3

2-]

(b) Fe(OH)2(s) Fe2+(aq) + 2OH-(aq) Ksp = [Fe2+][OH-]2

(c) Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq) Ksp = [Ca2+]3[PO4

3-]2

19-52

Sample Problem 19.5

(d) Ag2S(s) 2Ag+(aq) + S2-(aq) S2-(aq) + H2O(l) → HS-(aq) + OH-(aq)

Ksp = [Ag+]2[HS-][OH-]

Ag2S(s) + H2O(l) 2Ag+(aq) + HS-(aq) + OH-(aq)

19-53

Table 19.2 Solubility-Product Constants (Ksp) of Selected Ionic

Compounds at 25°C

Name, Formula Ksp

Aluminum hydroxide, Al(OH)3

Cobalt(II) carbonate, CoCO3

Iron(II) hydroxide, Fe(OH)2

Lead(II) fluoride, PbF2

Lead(II) sulfate, PbSO4

Silver sulfide, Ag2S

Zinc iodate, Zn(IO3)2

3x10-34

1.0x10-10

4.1x10-15

3.6x10-8

1.6x10-8

4.7x10-29

8x10-48

Mercury(I) iodide, Hg2I2

3.9x10-6

19-54

Sample Problem 19.6 Determining Ksp from Solubility

PROBLEM: (a) Lead(II) sulfate (PbSO4) is a key component in lead-acid car batteries. Its solubility in water at 25°C is 4.25x10-3 g/100 mL solution. What is the Ksp of PbSO4?

(b) When lead(II) fluoride (PbF2) is shaken with pure water at 25°C, the solubility is found to be 0.64 g/L. Calculate the Ksp of PbF2.

PLAN: We write the dissolution equation and the ion-product expression for each compound. This tells us the number of moles of each ion formed. We use the molar mass to convert the solubility of the compound to molar solubility (molarity), then use it to find the molarity of each ion, which we can substitute into the Ksp expression.

19-55

Sample Problem 19.6

Ksp = [Pb2+][SO42-]

Ksp = [Pb2+][SO42-] = (1.40x10-4)2

SOLUTION:

= 1.96x10-8

(a) PbSO4(s) Pb2+(aq) + SO42-(aq)

Converting from g/mL to mol/L:

4.25x10-3g PbSO4

100 mL solnx 1000 mL

1 Lx 1 mol PbSO4

303.3 g PbSO4 = 1.40x10-4 mol/L PbSO4

Each mol of PbSO4 produces 1 mol of Pb2+ and 1 mol of SO42-, so

[Pb2+] = [SO42-] = 1.40x10-4 mol/L

19-56

Sample Problem 19.6

Ksp = [Pb2+][F-]2

Ksp = [Pb2+][F-]2 = (2.6x10-3)(5.2x10-3)2 = 7.0x10-8

(b) PbF2(s) Pb2+(aq) + F-(aq)

Converting from g/L to mol/L:

0.64 g PbF2

1 L solnx 1 mol PbF2

245.2 g PbF2 = 2.6x10-3 mol/L PbF2

Each mol of PbF2 produces 1 mol of Pb2+ and 2 mol of F-, so [Pb2+] = 2.6x10-3 mol/L and [F-] = 2(2.6x10-3) = 5.2x10-3 mol/L

19-57

Sample Problem 19.7 Determining Solubility from Ksp

PROBLEM: Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH)2 are used in industry as a strong, inexpensive base. Calculate the molar solubility of Ca(OH)2 in water if the Ksp is 6.5x10-6.

PLAN: We write the dissolution equation and the expression for Ksp. We know the value of Ksp, so we set up a reaction table that expresses [Ca2+] and [OH-] in terms of S, the molar solubility. We then substitute these expressions into the Ksp expression and solve for S.

SOLUTION:

Ksp = [Ca2+][OH-]2 = 6.5x10-6 Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)

19-58

Sample Problem 19.7

-Initial 0 0

Change - +S + 2S

Ksp = [Ca2+][OH-]2 = (S)(2S)2 = 4S3 = 6.5x10-6

= 1.2x10-2

mol/L

Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)Concentration (mol/L)

Equilibrium - S 2S

S = =3 Ksp4 3 6.5x10-6

4

19-59

Table 19.3 Relationship Between Ksp and Solubility at 25°C

No. of Ions Formula Cation/Anion Ksp Solubility (mol/L)

2 MgCO3 1/1 3.5x10-8 1.9x10-4

2 PbSO4 1/1 1.6x10-8 1.3x10-4

2 BaCrO4 1/1 2.1x10-10 1.4x10-5

3 Ca(OH)2 1/2 6.5x10-6 1.2x10-2

3 BaF2 1/2 1.5x10-6 7.2x10-3

3 CaF2 1/2 3.2x10-11 2.0x10-4

3 Ag2CrO4 2/1 2.6x10-12 8.7x10-5

The higher the Ksp value, the greater the solubility, as long as we compare compounds that have the same total number of ions in their formulas.

19-60

Figure 19.12 The effect of a common ion on solubility.

PbCrO4(s) Pb2+(aq) + CrO42-(aq)

If Na2CrO4 solution is added to a saturated solution of PbCrO4, it provides the common ion CrO4

2-, causing the equilibrium to shift to the left. Solubility decreases and solid PbCrO4 precipitates.

19-61

Sample Problem 19.8 Calculating the Effect of a Common Ion on Solubility

PROBLEM: In Sample Problem 19.7, we calculated the solubility of Ca(OH)2 in water. What is its solubility in 0.10 mol/L Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10-6.

PLAN: The addition of Ca2+, an ion common to both solutions, should lower the solubility of Ca(OH)2. We write the equation and Ksp expression for the dissolution and set up a reaction table in terms of S, the molar solubility of Ca(OH)2. We make the assumption that S is small relative to [Ca2+]init because Ksp is low. We can then solve for S and check the assumption.

SOLUTION:

Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Ksp = [Ca2+][OH-]2

19-62

Sample Problem 19.8

Change - +S + 2S

Ksp = [Ca2+][OH-]2 = 6.5x10-6 ≈ (0.10)(2S)2 = (0.10)(4S2)

= 4.0x10-3 mol/L

Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)Concentration (mol/L)

Equilibrium - 0.10 + S 2S

[Ca2+]init = 0.10 mol/L because Ca(NO3)2 is a soluble salt, and dissociates completely in solution.

4S2 ≈ 6.5x10-6

0.10so S ≈6.5x10-5

4

Checking the assumption:4.0x10-3 mol/L

0.10 mol/Lx 100 = 4.0% < 5%

-Initial 0.10 0

19-63

Effect of pH on Solubility

Changes in pH affects the solubility of many slightly soluble ionic compounds.

The addition of H3O+ will increase the solubility of a salt that contains the anion of a weak acid.

CaCO3(s) Ca2+(aq) + CO32-(aq)

CO32-(aq) + H3O+(aq) → HCO3

-(aq) + H2O(l)

HCO3-(aq) + H3O+(aq) → [H2CO3(aq)] + H2O(l) → CO2(g) + 2H2O(l)

The net effect of adding H3O+ to CaCO3 is the removal of CO3

2- ions, which causes an equilibrium shift to the right. More CaCO3 will dissolve.

19-64

Figure 19.13 Test for the presence of a carbonate.

When a carbonate mineral is treated with HCl, bubbles of CO2 form.

19-65

Sample Problem 19.9 Predicting the Effect on Solubility of Adding Strong Acid

PROBLEM: Write balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of each ionic compound:

PLAN: We write the balanced dissolution equation for each compound and note the anion. The anion of a weak acid reacts with H3O+, causing an increase in solubility.

SOLUTION:

(a) lead(II) bromide (b) copper(II) hydroxide (c) iron(II) sulfide

(a) PbBr2(s) Pb2+(aq) + 2Br-(aq)

Br- is the anion of HBr, a strong acid, so it does not react with H3O+. The addition of strong acid has no effect on its solubility.

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Sample Problem 19.9

(b) Cu(OH)2(s) Cu2+(aq) + 2OH-(aq)

OH- is the anion of H2O, a very weak acid, and is in fact a strong base. It will react with H3O+:

The addition of strong acid will cause an increase in solubility.

OH-(aq) + H3O+(aq) → 2H2O(l)

(c) FeS(s) Fe2+(aq) + S2-(aq)

S2- is the anion of HS-, a weak acid, and is a strong base. It will react completely with water to form HS- and OH-. Both these ions will react with added H3O+:

The addition of strong acid will cause an increase in solubility.

HS-(aq) + H3O+(aq) → H2S(aq) + H2O(l)OH-(aq) + H3O+(aq) → 2H2O(l)

19-67

Figure 19.14 Limestone cave in Nerja, Málaga, Spain.

Limestone is mostly CaCO3 (Ksp = 3.3x10-9).

CO2(g) CO2(aq) H2O

CO2(aq) + 2H2O(l) H3O+(aq) + HCO3-(aq)

Ground water rich in CO2 trickles over CaCO3, causing it to dissolve. This gradually carves out a cave.

CaCO3(s) + CO2(aq) + H2O(l) Ca2+(aq) + 2HCO3-(aq)

Water containing HCO3- and Ca2+ ions

drips from the cave ceiling. The air has a lower P than the soil, causing CO2 to come out of solution. A shift in equilibrium results in the precipitation of CaCO3 to form stalagmites and stalactites.

CO2

19-68

Predicting the Formation of a Precipitate

For a saturated solution of a slightly soluble ionic salt, Qsp = Ksp.

When two solutions containing the ions of slightly soluble salts are mixed,

If Qsp = Ksp, the solution is saturated and no change will occur.If Qsp > Ksp, a precipitate will form until the remaining solution is saturated.

If Qsp =< Ksp, no precipitate will form because the solution is unsaturated.

19-69

Sample Problem 19.10 Predicting Whether a Precipitate Will Form

PROBLEM: A common laboratory method for preparing a precipitate is to mix solutions containing the component ions. Does a precipitate form when 0.100 L of 0.30 mol/L Ca(NO3)2 is mixed with 0.200 L of 0.060 mol/L NaF?

PLAN: First we need to decide which slightly soluble salt could form, look up its Ksp value in Appendix C, and write the dissolution equation and Ksp expression. We find the initial ion concentrations from the given volumes and molarities of the two solutions, calculate the value for Qsp and compare it to Ksp.

SOLUTION:

The ions present are Ca2+, NO3-, Na+, and F-. All Na+ and NO3

- salts are soluble, so the only possible precipitate is CaF2 (Ksp = 3.2x10-11).

CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = [Ca2+][F-]2

19-70

Sample Problem 19.10

Ca(NO3)2 and NaF are soluble, and dissociate completely in solution.

We need to calculate [Ca2+] and [F-] in the final solution.

Amount (mol) of Ca2+ = 0.030 mol/L Ca2+ x 0.100 L = 0.030 mol Ca2+.

[Ca2+]init =0.030 mol Ca2+

0.100 L + 0.200 L= 0.10 mol/L Ca2+

Amount (mol) of F- = 0.060 mol/L F- x 0.200 L = 0.012 mol F-.

[F-]init =0.012 mol F-

0.100 L + 0.200 L= 0.040 mol/L F-

Qsp = [Ca2+]init[F-]2init = (0.10)(0.040)2 = 1.6x10-4

Since Qsp > Ksp, CaF2 will precipitate until Qsp = 3.2x10-11.

19-71

Sample Problem 19.11 Using Molecular Scenes to Predict Whether a Precipitate Will Form

PROBLEM: These four scenes represent solutions of silver (gray) and carbonate (black and red) ions above solid silver carbonate. (The solid, other ions, and water are not shown.)

(a) Which scene best represents the solution in equilibrium with the solid?

(b) In which, if any, other scene(s) will additional solid silver carbonate form?

(c) Explain how, if at all, addition of a small volume of concentrated strong acid affects the [Ag+] in scene 4 and the mass of solid present.

19-72


PLAN: We need to determine the ratio of the different types of ion in each solution. A saturated solution of Ag2CO3 should have 2Ag+ ions for every 1 CO3

2- ion. For (b) we need to compare Qsp to Ksp. For (c) we recall that CO3

2- reacts with H3O+.

SOLUTION:

First we determine the Ag+/CO32- ratios for each scene.

Scene 1: 2/4 or 1/2 Scene 2: 3/3 or 1/1Scene 3: 4/2 or 2/1 Scene 4: 3/4

(a) Scene 3 is the only one that has an Ag+/CO32- ratio of 2/1, so this

scene represents the solution in equilibrium with the solid.

19-73

(b) We use the ion count for each solution to determine Qsp for each one. Since Scene 3 is at equilibrium, its Qsp value = Ksp.


Scene 1: Qsp = (2)2(4) = 16

Scene 3: Qsp = (4)2(2) = 32

Scene 2: Qsp = (3)2(3) = 27

Scene 4: Qsp = (3)2(4) = 36

Scene 4 is the only one that has Qsp > Ksp, so a precipitate forms in this solution.

(c) Ag2CO3(s) 2Ag+(aq) + CO32-(aq)

CO32-(aq) + 2H3O+(aq) → [H2CO3(aq)] + 2H2O(l) → 3H2O(l) + CO2(g)

The CO2 leaves as a gas, so adding H3O+ decreases the [CO32-] in

solution, causing more Ag2CO3 to dissolve.

[Ag+] increases and the mass of Ag2CO3 decreases.

19-74

Selective Precipitation

Selective precipitation is used to separate a solution containing a mixture of ions.

A precipitating ion is added to the solution until the Qsp of the more soluble compound is almost equal to its Ksp.

The less soluble compound will precipitate in as large a quantity as possible, leaving behind the ion of the more soluble compound.

19-75

Sample Problem 19.12 Separating Ions by Selective Precipitation

PROBLEM: A solution consists of 0.20 mol/L MgCl2 and 0.10 mol/L CuCl2. Calculate the [OH-] that would separate the metal ions as their hydroxides. Ksp of Mg(OH)2= is 6.3x10-10; Ksp of Cu(OH)2 is 2.2x10-20.

PLAN: Both compounds have 1/2 ratios of cation/anion, so we can compare their solubilities by comparing their Ksp values. Mg(OH)2 is 1010 times more soluble than Cu(OH)2, so Cu(OH)2 will precipitate first. We write the dissolution equations and Ksp expressions. Using the given cation concentrations, we solve for the [OH-] that gives a saturated solution of Mg(OH)2. Then we calculate the [Cu2+] remaining to see if the separation was successful.

19-76


SOLUTION:

Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) Ksp = [Mg2+][OH-]2 = 6.3x10-10

Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) Ksp = [Cu2+][OH-]2 = 2.2x10-20

Ksp

[Mg2+][OH-] = = 6.3x10-10

0.20= 5.6x10-5 mol/L

This is the maximum [OH-] that will not precipitate Mg2+ ion.

Calculating the [Cu2+] remaining in solution with this [OH-]

Ksp

[OH-]2[Cu2+] = =

2.2x10-20

(5.6x10-5)2= 7.0x10-12 mol/L

Since the initial [Cu2+] is 0.10 mol/L, virtually all the Cu2+ ion is precipitated.

19-77

Figure B19.1 Formation of acidic precipitation.

Chemical Connections

Since pH affects the solubility of many slightly soluble ionic compounds, acid rain has far-reaching effects on many aspects of our environment.

19-78

Figure 19.15 Cr(NH3)63+, a typical complex ion.

A complex ion consists of a central metal ion covalently bonded to two or more anions or molecules, called ligands.

19-79

Figure 19.16 The stepwise exchange of NH3 for H2O in M(H2O)42+.

The overall formation constant is given by

Kf = [M(NH3)4

2+]

[M(H2O)42+][NH3]4

19-80

Table 19.4 Formation Constants (Kf) of Some Complex Ions at 25°C

19-81

Sample Problem 19.13 Calculating the Concentration of a Complex Ion

PROBLEM: An industrial chemist converts Zn(H2O)42+ to the more

stable Zn(NH3)42+ by mixing 50.0 L of 0.0020 mol/L

Zn(H2O)42+ and 25.0 L of 0.15 mol/L NH3. What is the

final [Zn(H2O)42+] at equilibrium? Kf of Zn(NH3)4

2+ is 7.8x108.PLAN: We write the reaction equation and the Kf expression, and

use a reaction table to calculate equilibrium concentrations. To set up the table, we must first find [Zn(H2O)4

2+]init and [NH3]init using the given volumes and molarities. With a large excess of NH3 and a high Kf, we assume that almost all the Zn(H2O)4

2+ is converted to Zn(NH3)42+.

SOLUTION:

Zn(H2O)42+(aq) + 4NH3(aq) Zn(NH3)4

2+(aq) + 4H2O(l)

Kf =[Zn(NH3)4

2+]

[Zn(H2O)42+][NH3]4

19-82


[Zn(H2O)42+]initial = = 1.3x10-3

mol/L

50.0 L x 0.0020 mol/L

50.0 L + 25.0 L

[NH3]initial = = 5.0x10-2

mol/L

25.0 L x 0.15 mol/L

50.0 L + 25.0 L

Initial 1.3x10-3 5.0x10-2 0 -

Change ~(-1.3x10-3) ~(-5.2x10-3) ~(+1.3x10-3) -

Zn(H2O)42+(aq) + 4NH3(aq) Zn(NH3)4

2+(aq) + 4H2O(l)Concentration (mol/L)

Equilibrium x 4.5x10-2 1.3x10-3 -

4 mol of NH3 is needed per mol of Zn(H2O4)2+, so[NH3]reacted = 4(1.3x10-3 mol/L) = 5.2x10-3 mol/L and [Zn(NH3)4

2+] ≈ 1.3x10-3 mol/L

19-83

Kf =[Zn(NH3)4

2+]

[Zn(H2O)42+][NH3]4

= 7.8x108 =(1.3x10-3)

x(4.5x10-2)4

x = [Zn(H2O)42+ = 4.1x10-7

mol/L


19-84

Sample Problem 19.14 Calculating the Effect of Complex-Ion Formation on Solubility

PROBLEM: In black-and-white film developing, excess AgBr is removed from the film negative by “hypo”, an aqueous solution of sodium thiosulfate (Na2S2O3), which forms the complex ion Ag(S2O3)2

3-. Calculate the solubility of AgBr in (a) H2O; (b) 1.0 mol/L hypo. Kf of Ag(S2O3)2

3- is 4.7x1013 and Ksp AgBr is 5.0x10-13.

PLAN: After writing the equation and the Ksp expression, we use the given Ksp value to solve for S, the molar solubility of AgBr. For (b) we note that AgBr forms a complex ion with S2O3

2-, which shifts the equilibrium and dissolves more AgBr. We write an overall equation for the process and set up a reaction table to solve for S.

19-85


SOLUTION:

S = [AgBr]dissolved = [Ag+] = [Br-]

Ksp = [Ag+][Br-] = S2 = 5.0x10-13

(b) Write the overall equation:

(a) AgBr(s) Ag+(aq) + Br-(aq) Ksp = [Ag+][Br-] = 5.0x10-13

AgBr(s) + 2S2O32-(aq) Br-(aq) + Ag(S2O3)2

3-(aq)

S = 7.1x10-7 mol/L

AgBr(s) Ag+(aq) + Br-(aq)

Ag+(aq) + 2S2O32-(aq) Ag(S2O3)2

3-(aq)

Koverall = Ksp x Kf = [Br-][Ag(S2O3)2

3-]

[S2O32-]2

= (5.0x10-13)(4.7x1013) = 24

19-86


Initial - 1.0 0 0

Change - -2S +S +S

AgBr(s) + 2S2O32-(aq) Br-(aq) + Ag(S2O3)2

3-(aq)Concentration (mol/L)

Equilibrium - 1.0 - 2S S S

Koverall =S2

(1.0 - 2S)2= 24

S = [Ag(S2O3)23-] = 0.45

mol/L

S

1.0 - 2S= = 4.924so

S = 4.9 mol/L – 0.9S and 10.9S = 4.9 mol/L

19-87

Figure 19.17 The amphoteric behavior of aluminum hydroxide.

When solid Al(OH)3 is treated with H3O+ (left) or with OH- (right), it dissolves as a result of the formation of soluble complex ions.

19-1 chem 1b: general chemistry chapter 19: ionic equilibria in aqueous systems instructor: dr....

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