Calculate the degree of hydrolysis of $0.2 (M)$ sodium acetate solution.
(Hydrolysis constant of sodium acetate $= 5.6 \times 10^{- 10}$ and ionic product of $H_{2} O = 10^{- 14}$ at $25^{\circ} C$ )

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Solution

Degree of hydrolysis $h = \sqrt{\frac{K_{h}}{c}}$
given $K_{h} = 5.6 \times 10^{- 10}$
$c = 0.2 M$
Hence, $h = \sqrt{\frac{5.6 \times 10^{- 10}}{0.2}}$
$= 5.29 \times 10^{- 5}$

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Calculate the degree of hydrolysis of 0.2(M) sodium acetate solution.(Hydrolysis constant of sodium acetate =5.6×10−10 and ionic product of H2O=10−14 at 25∘C)

Degree of hydrolysis h=√Khcgiven Kh=5.6×10−10c=0.2 MHence, h=√5.6×10−100.2=5.29×10−5

Calculate the degree of hydrolysis of $0.2 (M)$ sodium acetate solution.
(Hydrolysis constant of sodium acetate $= 5.6 \times 10^{- 10}$ and ionic product of $H_{2} O = 10^{- 14}$ at $25^{\circ} C$ )

Degree of hydrolysis $h = \sqrt{\frac{K_{h}}{c}}$
given $K_{h} = 5.6 \times 10^{- 10}$
$c = 0.2 M$
Hence, $h = \sqrt{\frac{5.6 \times 10^{- 10}}{0.2}}$
$= 5.29 \times 10^{- 5}$