chemistry. ionic equilibrium-ii session objectives
TRANSCRIPT
Chemistry
Ionic equilibrium-II
Session Objectives
1. pH of weak acids
2. pH of mixture of two strong acids
3. pH of mixture of strong and weak acids
4. Dissociation of polybasic acids
5. pH of mixture of two weak acids
Session Objectives
For mixture of two strong acids
Let us consider a mixture containing 200 ml 0.1 M HCl and 500 ml 0.2 M H2SO4.Since both are strong electrolytes,
HCl H Cl
200 × 0.1×10–3 0.02 mole
= 0.02 mole2
2 4 4H SO 2H SO
500 × 0.2 × 10–3 2 × 0.1 mole
= 0.1 mole
For mixture of two strong acids
Total0.02 0.2
[H ] 0.3140.7
+
TotalpH = -log H = -log 0.314 = 0.503
Note: For mixture of any two acids or bases, degree of ionization of water is taken negligible because of common ion effect.
Total ion concentration,
Mixture of strong and weak acids
Let us consider a mixture of strong acid(HA) and a weak acid (HX) of concentration C1 and C2 respectively.
1 1 1
HA H AC C C
For weak acid,
2 2 2
HX H XC 1 C C
Dissociation constant of weak acid,
1 2 2Totala
2
C C C[H ] [X ]K
[HX] C 1
Now, for strong acid
Mixture of strong acid and weak acid
1 2 2Totala
2
C C C[H ] [X ]K
[HX] C 1
1 2 pH = -log(C + C )
where, degree of dissociation of weak acid.
Question
Illustrative example 1
A solution contains 0.10 M H2S and0.3 M HCl. Calculate the concentration of [HS–] and [S–2] ions in the solution.For H2S,Ka1 = 1.0 x 10-7 Ka2 = 1.3 x 10-13
Solution:
0.30.3HCl H Cl
H2S H+ + HS–
1
7a
2
[H ][HS ]K 1 10
[H S]
Solution
2
213
a[H ][S ]
K 1.3 10[HS ]
2[H ] coming from H S is negligibly small due to
common ion effect
[H ] 0.3M.
7 0.3 [HS ]10
0.1
8[HS ] 3.33 10
HS– H+ + S–2
Solution
Considering [HS–] dissociates to a very small extent
213
80.3 [S ]
1.3 103.33 10
212 214.33 10
[S ] 14.43 100.3
201.443 10
Dissociation of polybasic acids
Acids giving more than one hydrogen ion per molecule are ‘polybasic’ or ‘polyprotic’ acids. Examples are H2C2O4, H2CO3, H2S, H3PO4, H3AsO4, etc.
H HS 2H S1
2
H HSK
H S
HS H S 2
H SK
HS
These dissociate in stages . For example,
Dissociation of polybasic acids
Normally K2 << K1.
To calculate hydrogen ion concentration, only the first step should be considered as the H+ obtained from successive dissociation can be neglected, but to calculate the concentration of
then both the equilibria have to be considered.
2[HS ],[S ] etc.
pH of mixture of two weak acids
Let HA and HB are two weak acids.
HA H A
1 1 1 1 1 1C (1 ) C C
HB H B
2 2 2 2 2 2C (1 ) C C
Total1
[H ] [A ]Ka
[HA]
1 1 2 2 2 2
2 2
(C C )C(2)
C (1 )
Total2
[H ] [B ]Ka
[HB]
1 1 2 2 1 1
1 1
(C C )C(1)
C (1 )
pH of mixture of two weak acids
1
2
a 1
a 2
K
K
1 2, 1
TotalSolving this equation we can calculate [H ]
and hence pH
Dividing (1) by (2)
Question
Illustrative example 2A solution is prepared by mixing 0.2MHCOOH with 0.5 M CH3 COOH.Given KaCH3COOH=1.8 x 10–5 , KaHCOOH =2.1x10-4
Calculate [HCOO–] , [CH3COO–] and pH of the solution.Solution:
10.2(1 )HCOOH
1 2(0.2 0.5 )H
10.2HCOO
2
30.5(1 )
CH COOH 1 2(0.2 0.5 )
H
2
30.5
CH COO
4 1 2 11
1
(0.2 0.5 )K 2.1 10 ......(1)
(1 )
Solution
5 1 2 22
2
(0.2 0.5 )K 1.8 10
(1 )
......(2)
1 2, 1 4
15
2
2.1 10
1.8 10
11 2
211.67 11.67
52 2 2
22
1.8 10 (0.2 11.67 0.5 )
2.834
From (2)
Solution
53
21.8 10
2.52 102.834
31
2
11.67 2.5 10
2.94 10
total 1 2
3 3
H (0.2 0.5 )
(5.88 10 1.26 10 )
37.14 10
Solution
mixpH 3 log7.14 2.15
31HCOO 0.2 5.88 10
33 2CH COO 0.5 1.25 10
Hydrolysis of Salts
A. Hydrolysis of a salt of weak acid and strong base
The hydrolysis reaction is
3 2 3
ChC Ch ChCH COO H O CH COOH OH
At eqm.
where C = concentration of salt h = degree of hydrolysis.
4 2 4NH Cl H O NH OH H Cl
2HCOONa H O HCOOH OH Na
Hydrolysis of Salts
23h
3
CH COOH OH ChK
(1 h)CH COO
2hK ch h 1
2wh
a
KK Ch
K 2 w w
a a
K Kh or h
K .C K .C
w
a
OH Ch
K . C
K
Hydrolysis constant
Hydrolysis of Salts
w a
1 1 1log OH log K log K log C
2 2 2
w a1
pOH pK log K logC2
w a1 1 1
pH pK pK logC2 2 2
Hydrolysis of Salts
4 4For NH Cl NH Cl
4 2 4NH H O NH OH H
2wh
b
KSimilarly, K Ch
K
where Kb = Dissociation constant of weak base.
w
b
Kh
K C
w
b
KH Ch C
K
B. Salt hydrolysis of strong acid and weak base
w b
1 1 1pH pK pK logC
2 2 2
Hydrolysis of Salts
C. Hydrolysis for a salt of weak acid and weak base
3 4 3 4CH COONH CH COO NH
3 2 3CH COO H O CH COOH OH
C 1 h Ch Ch
At eqm.
4 2 4NH H O NH OH H
C 1 h Ch Ch
At eqm.
Hydrolysis of Salts
3 4h
3 4
2 2w
22a b
CH COOH NH OHK
CH COO NH
K c h
K K c 1 h
w
a b
Kh=
1- h K ×K
Hydrolysis of Salts
w a b
1 1 1pH pK pK pK
2 2 2
w aa
b
1 h K KH K
h K
3 3CH COOH H CH COOC 1 h Ch Ch
For pH,
Now, to calculate the pH
3
a3
H CH COO H ChK
CH COOH C 1 h
Questions
Illustrative example 3Calculate the pH at the equivalence point of the titration between 0.1 M CH3COOH (50 ml) with 0.05 M NaOH . Ka(CH3COOH) = 1.8 × 10–5.
At the equivalence point,
3 3 23 3
CH COOH NaOH CH COONa H O
50 0.1 10 V 0.05 10
Let V ml NaOH is required to reach the equivalence point.
1 1 2 23 3
N V N V
50 0.1 10 V 0.05 10
Solution:
At the equivalence point,
Solution
V = 100 ml
3 2 3CH COO H O CH COOH OHC(1 h) Ch Ch
2wh
a
KK Ch
K
w
a
Kh
K C
[OH ] C . h
30.1 50
CH COONa 0.033150
Solution
w aw w K KK K[H ]
Ch C[OH ]
14 5
2
10 1.8 10
3.3 10
pH log[H ] 9 log 2.33 8.63
18
195.45 10
2.33 10
Illustrative example 4
Calculate the percentage hydrolysis of decinormal solution of ammonium acetate, given Ka = 1.75 × 10–5, Kb = 1.80 × 10–5 and Kw = 1 × 10–14. What will be the change in degree of hydrolysis when 2 L of water is added to 1 L of the above solution?
Since CH3COONH4 is a salt of weak acid and weak base
2w
h 2a b
K hK
K K (1 h)
Solution:
Solution
14
5 51 10
1.75 10 1.8 10
35.63 10
3h 5.63 10 [h 1]
Since degree of hydrolysis is not affected by the concentration in this case. So on dilution there will be no change in degree of hydrolysis.
w
a b
Kh1 h K K
Illustrative example 5
Hydrolysis constant of Zn+2 is 1 × 10–9
(a) Calculate pH of a 0.001 M solution of ZnCl2.
(b) What is the basic dissociation constant of Zn(OH)+?
223 8310 (1 h) 10 h10 h
Zn H O Zn(OH) H
9 3 2hK 1 10 10 h h 1
Solution:
Solution
93
310
h 1010
3 6H 10 h 10
pH 6
23 33 10 1010 h
Zn(OH) Zn OH
2
b
Zn OHK
Zn(OH)
11
610
10
510
Illustrative example 6
When 0.20 M acetic acid is neutralised with 0.20 M NaOH in 0.50 litre of water, the resulting solution is slightly alkaline, calculate the pH of the resulting solution (Ka for acetic acid = 1.8 x 10-5)
3 3 2CH COOH NaOH CH COONa H O
moles 0.2 0.5 0.2 0.5 0.2 0.5
0.1(1 h) 0.1h 0.1h
3 2 3CH COO H O CH COOH OH
Solution:
Solution
22w
ha
K 0.1hK 0.1h
K 1 h
h 1
144
5
10h 0.745 10
1.8 10 0.1
OH 0.1h pOH 5.128
pH 8.872
Illustrative example 7Calcium lactate is a salt of a weak organic acid and represented as Ca(Lac)2. A saturated solution of Ca (Lac)2 contains 0.13 mol of this salt in 0.50 litre solution. The pOH of this solution is 5.60. Assuming a complete dissociation of the salt, calculate Ka of lactic acid.
2 12Ca(Lac) Ca 2Lac
0.13mol 2 0.13 mol.
0.26Lac 0.52 M
0.5
Solution:
Solution
0.52(1 h) 0.52h 0.52h
6
6
OH 0.52h 2.51 10
h 4.83 10
2Lac H O LacH OH
2wh
a
KK 0.52h h 1
K
14
a 26
10K
0.52 4.83 10
48.24 10
Class exercise
Class exercise 1
The hydrolysis constant for FeCl2 will be
wh
b
K(a) K =
K
2w
wb
K(b) K =
K
2w
h 2b
K(c) K =
Kb
h 2w
K(d) K =
K
FeCl2 is the salt of strong acid and weak base.+2 -
2Since FeCl Fe + 2Cl+2 +
2 2Fe + 2H O Fe(OH) + 2H
Solution:
Solution
+ 22
h +2[Fe(OH) ] [H ]
K =[Fe ]
+2 -2Fe(OH) Fe +2OH
+2 - 2 2+ + 2
b + 22 2
[Fe ][OH ] [Fe ][OH] [H ]K = = ×
[Fe(OH) ] [Fe(OH) ] [H ]
2w
hb
KK =
K
Hence, the answer is (b)
Class exercise 2
pH of a solution produced when an aqueous solution of pH 6 is mixed with an equal volume of an aqueous solution of pH 3, will be
(a) 4.5 (b) 4.3
(c) 4.0 (d) 3.3
pH = 6
+ -6[H ]=10 M
pH = 3
-6 -3+
Total10 +10
[H ] =2
Solution:
Solution
-3-41.001×10
= =5.005×102
+Total pH = – log [H ]
= 4 – log 5.005
= 3.3
Hence, the answer is (d).
Class exercise 3
Which one of the following is true forany diprotic acid, H2X?
(a) Ka2 > Ka1(b) Ka1 > Ka2
(c) Ka1 = Ka2 21
1(d) Ka =
Ka
H2X being a diprotic acid,
+ -2 1
- + -22
H X H + HX Ka
HX H + X Ka
Due to the ‘common ion effect’ dissociation of HX– will be less.
2 1 Ka > Ka
Solution:
Class exercise 4
Ka (CH3COOH) = 1.7 × 10–5 and [H+] = 3.4 × 10–4. Then initial concentrations of CH3COOH is
(a) 3.4 × 10–4 (b) 6.8 × 10–3
(c) 3.4 × 10–3 (d) 6.8 × 10–2
+ -3 3CH COOH H + CH COO
C(1– ) C C
2CKa=
1-2C
Solution:
Solution
+ -4H ]=3.4×10 =[ C
–5 –4 1.7 10 = 3.4 × 10
-5-2
-41.7×10
=5×103.4×10
-4-2 -3
-23.4×10
=0.68×10 =6.8×105×10
C
Hence, the answer is (b).
Class exercise 5
0.001 M HCl is mixed with 0.01 M HCOOH at 25° C. If Ka HCOOH = 1.7 × 10–4, find the pH of the resulting solution.
+ -HCl H + Cl0.001 0.001 0.001
+ -HCOOH H + HCOO
–2
–2 –2 -2
Initial mole 10
At eqm. 10 (1– ) 10 +10
+
TotalH = 2 310 10
Solution:
Solution
+ -
TotalH HCOO
Ka=HCOOH
-2 -3
-2
10 +10
10 1-2 3
aK 10 10
–4 –3 –2 2 –31.7 × 10 = 10 [ 10 << 10 ]
+ -2 -3H = =10 ×.17=1.7×10C–1 = 1.7 × 10
+ -3 -3 -3
TotalH = 1.7×10 +10 =2.7×10
pH = 3 – log 2.7 = 2.57
Class exercise 6What is the percentage hydrolysis of NaCN solution when the
Ka HCN = 1.3 × 10–9,K2 = 1 × 10–14?
(a) 2.48 (b) 5.26
(c) 9.6 (d) 8.2
N80
+ -NaCN Na + CN- -
2CN + H O HCN + OH
11- h
80
h80
h80
Solution:
Solution
2
2w
ha
hK h80K = =
1K 801- h80
-14 2
-1910 h
801.3×10
-5-480×10
h= = 6.15×101.3
= 2.48 × 10–2
–2 2Percentage hydrolysis = 2.48 × 10 × 10 = 2.48
Class exercise 7
Calculate the pH of a solution obtained by mixing 100 ml of 0.1 M HCl with 9.9 ml of 1 M NaOH.
2-3 -3
-4 -4
HCl + NaOH NaCl + H O
100×0.1×10 9.9×1×10
100×10 99×10
Acid remaining = 1 × 10–4 moles
Solution:
pH = –log[H+] = 4 – log 9.09
= 3.04
-4+ 3 -41×10
H = ×10 =9.09×10109.9
Class exercise 8Calculate the percentage hydrolysis of 3 × 10–3 M aqueous solution of NaOCN (Ka HCON = 3.33 × 10–4 M).
+ -NaOCN Na + OCN
- -2
-3 -3 -3OCN + H O HOCN + OH
3×10 1- h 3×10 h 3×10 h
-3 2-3 2w
ha
K 3×10 hK = = 3×10 h
K 1- h
Solution:
Solution
-14-8 -4
-4 -310
h= = 1×10 =103.33×10 ×3×10
-4 2Percentage hydrolysis = 10 ×10 = 0.01
Thank you