Salt hydrolysis

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## Salt hydrolysis-

### Q 1) Calculate for 0.01 N solution of CH_{3}COONa

### i) hydrolysis constant (ii) degree of hydrolysis (iii) pH

### (K_{a} = 1.9 x 10^{-5})

### Solution )

### i)

Sodium acetate is a salt of strong base and weak acid.

K_{w} = 10^{-14}

K_{h} =K_{w} / K_{a}

K_{h} = 10^{-14} /1.9 x 10^{-5}

### Hydrolysis constant ‘K_{h} ‘= 5.26 x 10 ^{-10} Ans.

#### ii)

Salt hydrolysis-

CH_{3}COO^{–} + H_{2}O ** ⇌ **CH

_{3}COOH + OH

^{–}

C (1-h) Ch

C = 0.01, K_{h} = 5.26 x 10 ^{-10}

K_{h} = Ch^{2}

h = **√ K _{h} / C
**

h = **√ 5.26 x 10 ^{-10} / 0.01
**

h = **√ 526 x 10 ^{-10}
**

### h = 22.9 x 10 -5

### Degree of hydrolysis, h = 2.29 x 10^{-4} Ans.

### iii)

[OH^{–}] = Ch = 0.01 x 2.29 x 10 ^{-4} = 2.29 x 10^{-6} M

[H^{+}] = K_{w} / [OH^{–}] = 10^{-14} / 2.29 x 10^{-6}

[H^{+}] = 4.37 x 10 ^{-9} M

pH = -log [H^{+}]= – log (4.37 x 10 ^{-9})

= – (log 4.37 + log 10^{-9})

= – (log 4.37 -9 log 10)

= -log 4.37 + 9 log 10

= -0.6405 + 9= 8.36

### pH = 8.36 Ans.

### Alternate method for calculation of pH-

CH3COONa is a salt of strong base and weak acid. Therefore ,

pH = 7 + 1/2 pK_{a} + 1/2 log C

K_{a} =1.9 x 10 ^{-5}, C = 0.01 M

pK_{a} = -log K_{a}= -log (1.9 x 10 ^{-5})

= -(log 1.9 + log 10 ^{-5})

= -(log 1.9 – 5 log 10 )

= – ( 0.2787 – 5 x 1)

= – ( – 4.7213)

pK_{a} = 4.7213

pH = 7 + (4.7213/2) + 1/2 log 0.01=7 + 2.36 + (-2/2)

pH= 9.36 -1.00

### pH = 8.36 Ans.

### Q 2) Calculate the % hydrolysis in 0.003 M aqueous solution of NaOCN . K_{a} for HOCN = 3.33 x 10 ^{-4}.

### Solution )

#### It is a salt of strong base and weak acid.

h = **√ K _{w}/(K_{a}. C)**

C = 0.003 M , K_{a} = 3.33 x 10 ^{-4} , K_{w} = 10 ^{-14}

h = **√ 10 ^{-14} / 3.33 x 10^{-4} x 0.003**

h = **√ 10 ^{-8} **

h = 10 ^{-4}

% of h = 10 ^{-4} x 100 = 10^{-2}

### % of h = 0.01 % Ans.

### Q 3) Calculate the pH of a 0.5 M aqueous NaCN solution . Given pK_{b} 0f CN ^{–} = 4.7

### Solution )

pK_{a} for HCN = 14 – 4.7= 9.3

NaCN is a salt of strong base and weak acid.Therefore,

[OH^{–}] = Ch

h = **√ K _{w}/(K_{a}. C)**

[OH^{–}] = C **√ K _{w}/(K_{a}. C)**

[OH^{–}] = **√ K _{w} .C / K_{a}**

Taking log and reverting the sign-

-log [OH^{–}] = 1/2 [-log K_{w} – log C + log K_{a}]

-log [OH^{–}] = pOH , -log K_{w} = pK_{w} , -log K_{a} = pK_{a}

pOH = 1/2 [pK_{w} -log C – pK_{a}]

= 1/2[14 – log 0.5 – 9.3]

= 1/2 [ 4.7 + 0.3010]

pOH = 2.5

pH = 14 – pOH = 14 – 2.5