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