A weak acid H X has the dissociation constant of 1 × 10-5 M. It forms a salt Na X on reaction with alkali. The degree of hydrolysis of 0.1 M solution of Na X is

Question

A weak acid H X has the dissociation constant of 1 × 10-5 M. It forms a salt Na X on reaction with alkali. The degree of hydrolysis of 0.1 M solution of Na X is (A) 0.0001 % (B) 0.01 % (C) 0.1 % (D) 0.15 %

Tardigrade Admin · Accepted Answer

For a salt of weak acid ( H X) and strong base ( NaOF We know, h=√(Kw/Ka . C)=√(10-14/1 × 10-5 × 0.1)=√10-8=10-4 Percentage degree of hydrolysis =10-4 × 100=10-2 i.e., 0.01 %

Q. A weak acid $H X$ has the dissociation constant of $1 \times 1 0^{- 5} M$ . It forms a salt $N a X$ on reaction with alkali. The degree of hydrolysis of $0.1 M$ solution of $N a X$ is

$0.0001%$

$0.01%$

$0.1%$

$0.15%$

For a salt of weak acid $(H X)$ and strong base $(N a OF$ We know, $h = \frac{K _{w}}{K _{a} . C} = \frac{1 0 ^{- 14}}{1 \times 1 0 ^{- 5} \times 0.1} = 1 0^{- 8} = 1 0^{- 4}$

Percentage degree of hydrolysis $= 1 0^{- 4} \times 100 = 1 0^{- 2}$ i.e., $0.01%$

Q. A weak acid HX has the dissociation constant of 1×10−5M. It forms a salt NaX on reaction with alkali. The degree of hydrolysis of 0.1M solution of NaX is

0.0001%

0.01%

0.1%

0.15%