Question

Calculate the degree of hydrolysis of 0.05 M solution of sodium acetate at 298 K. (Ka of acetic acid = 1.8 × 10–5)​

Answer 1

Thus the Degree of hydrolysis of 0.05 M Sodium acetate is 1.05 x 10^-4

Explanation:

Equilibrium reaction:

CH3COONa ----> Na+   + CH3COO -

Ka of acetic acid = 1.8 × 10–5

C = 0.05 M                     (Since Kw = 1 x 10^-14)

Degree of hydrolysis = √Kw / Kac

Degree of hydrolysis  = √ Kw / Kac  = √ 1 x 10^-14 / 1.8 x 10^-5 x 0.05

= √ 1 x 10^-14 + 5 / 0.09

= √ 0.1 x 10^-8 / 0.09

= √  1.11 x 10^-8

= 1.05 x 10^-4

Thus the Degree of hydrolysis of 0.05 M Sodium acetate is 1.05 x 10^-4

Answer 2

Given :

Concentration of sodium acetate , c = 0.05 M .

Temperature , t = 298 K .

$K_a=1.8\times 10^{-5}$ .

To Find :

The digree of hydrolysis of solution of sodium acetate .

Solution :

The chemical equation is :

                 $CH_3COO^- +H_2O \Longleftrightarrow CH_3COOH+OH^-$

At t = 0 s      0.1                                      0                0

At $t=t_{eqb}$     0.1 × (1 - h)                       0.1 × h        0.1 × h

$K_a=\dfrac{(0.1\times h)(0.1\times h)}{0.1\times (1-h)}\\\\K_a=\dfrac{(0.1\times h)(0.1\times h)}{0.1}\\\\1.8 \times 10^{-4}= h^2\\h=1.34\times 10^{-2}$

the degree of hydrolysis is $h=1.34\times 10^{-2}$ .

Hence , this is the required solution .

Learn More :

Equilibrium

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