The degree of hydrolysis and pH of $0.1 M$ sodium acetate solution are: Also calculate the concentration of all the ions at equilibrium. Hydrolysis constant of sodium acetate is $5.6 \times 10^{- 10}$ .

h = 7.5 x 10^-7 M, pH = 7.9

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h = 8.5 x 10^-6, pH = 6.9

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h = 8.5 x 10^-7, pH = 7.9

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h = 7.5 x 10^-6, pH = 7.9

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Solution

The correct option is D
h = 7.5 x 10^-6, pH = 7.9

When we have salts of strong base and weak acid,

$h$ = $\sqrt{\frac{K h}{c}}$

= $7.5 \times 10^{- 6}$

The hydrolysis of Sodium acetate happens as:

$C H_{3} C O O^{-} + H_{2} O ⇌ C H_{3} C O O H + O H^{-}$

$c (1 - h)$ $c h$ $c h$

$[O H^{-}]$ = $[C H 3 C O O H]$ = $c h$ = $7.5 \times 10^{- 7} M$

$[C H 3 C O O^{-}]$ = $c (1 - h)$ = $(1 - 7 .5 \times 10^{- 6}) \times 0.1 m o l L^{- 1}$ = $0.10000075 M$

$p H$ = $- L o g \frac{K w}{[O H^{-}]}$ = $8 - l o g 1.3$ = $7.9$

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Similar questions

0.1 M

5.6 \times 10^{- 10}

The degree of hydrolysis and pH of $0.1 M$ sodium acetate solution are:
Hydrolysis constant of sodium acetate is $5.6 \times 10^{- 10}$ .

10^{- 1} M

10^{- 3} M

C H_{3} C O O H [K_{a} = 2 \times 10^{- 5}]

[H^{+}]

C H_{3} C O O H .

h

0.1 M

298 K

K_{a} = 1.8 \times 10^{- 5}

H_{2} {C O}_{3}

H_{2} {C O}_{3} ⇌ H^{+} + H {C O}_{3}^{-}; K_{1} = 4.3 \times 10^{- 7}

H {C O}_{3}^{-} ⇌ H^{+} + {C O}_{3}^{2 -}; K_{2} = 5.6 \times 10^{- 11}

0.12 M

{N a}_{2} {C O}_{3}

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