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Chemistry

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Chemistry

Ionic equilibrium-II

Session Objectives

- pH of weak acids
- pH of mixture of two strong acids
- pH of mixture of strong and weak acids
- Dissociation of polybasic acids
- pH of mixture of two weak 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,

200 × 0.1×10–3 0.02 mole

= 0.02 mole

500 × 0.2 × 10–3 2 × 0.1 mole

= 0.1 mole

Total ion concentration,

Note: For mixture of any two acids or bases, degree of ionization of water is taken negligible because of common ion effect.

Let us consider a mixture of strong acid(HA) and a weak acid (HX) of concentration C1 and C2 respectively.

Now, for strong acid

For weak acid,

Dissociation constant of weak acid,

Question

H2S H+ + HS–

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 10-7Ka2 = 1.3 x 10-13

Solution:

HS– H+ + S–2

Considering [HS–] dissociates to a very small extent

Acids giving more than one hydrogen ion per molecule are ‘polybasic’ or ‘polyprotic’ acids.

Examples are H2C2O4, H2CO3, H2S, H3PO4, H3AsO4, etc.

These dissociate in stages . For example,

Normally K2 << K1.To calculate hydrogen ion concentration, only the first step should be consideredas the H+ obtained from successive dissociation can be neglected, but to calculate the concentration of

then both the equilibria have to be considered.

Let HA and HB are two weak acids.

Dividing (1) by (2)

Question

A 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:

From (2)

A. Hydrolysis of a salt of weak acid and strong base

The hydrolysis reaction is

At eqm.

where C = concentration of salt

h = degree of hydrolysis.

Hydrolysis constant

B. Salt hydrolysis of strong acid and weak base

where Kb = Dissociation constant of weak base.

C. Hydrolysis for a salt of weak acid and weak base

At eqm.

At eqm.

Now, to calculate the pH

For pH,

Questions

Calculate 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.

Solution:

At the equivalence point,

Let V ml NaOH is required to reach the equivalence point.

At the equivalence point,

V = 100 ml

Calculate the percentage hydrolysis of decinormal solution of ammonium acetate, given Ka = 1.75 × 10–5, Kb = 1.80 × 10–5and 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?

Solution:

Since CH3COONH4 is a salt of weak acid and weak base

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.

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)+?

Solution:

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)

Solution:

Calcium 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.

Solution:

Class exercise

The hydrolysis constant for FeCl2 will be

Solution:

FeCl2 is the salt of strong acid and weak base.

Hence, the answer is (b)

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

Solution:

pH = 6

pH = 3

= 4 – log 5.005

= 3.3

Hence, the answer is (d).

Which one of the following is true forany diprotic acid, H2X?

(a) Ka2 > Ka1(b) Ka1 > Ka2(c) Ka1 = Ka2

Solution:

H2X being a diprotic acid,

Due to the ‘common ion effect’ dissociation of HX–will be less.

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

Solution:

Hence, the answer is (b).

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.

Solution:

What 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

Solution:

= 2.48 × 10–2

Calculate the pH of a solution obtained by mixing 100 ml of 0.1 M HCl with 9.9 ml of 1 M NaOH.

Solution:

Acid remaining = 1 × 10–4 moles

pH = –log[H+] = 4 – log 9.09

= 3.04

Calculate the percentage hydrolysis of 3 × 10–3 M aqueous solution of NaOCN (Ka HCON = 3.33 × 10–4 M).

Solution:

Thank you