Ionization constants of weak acid HA and weak base BOH are 3-0 - 10-7 each at 298 K- The percent degree of hydrolysis of their salt at the dilution of 10 L is -A- 50B- 25C- 75D- 40

The correct option is B 25 k_H = K_w/K_a K_b = 1 × 0 × 10^ 14/(3 × 10^ 7)^2 = 1/9 h/1 h = √(K_H) = √(1/9)⇒ h = 0.25 ...

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Byju's Answer

Standard XII

Chemistry

Introduction and Degree of Hydrolysis

Ionization co...

Question

Ionization constants of weak acid HA and weak base BOH are $3.0 \times 10^{- 7}$ each at 298 K. The percent degree of hydrolysis of their salt at the dilution of 10 L is :

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Solution

The correct option is D
25

$k_{H} = \frac{K_{w}}{K_{a} K_{b}} = \frac{1 \times 0 \times 10^{- 14}}{(3 \times 10^{- 7})^{2}} = \frac{1}{9} \frac{h}{1 - h} = \sqrt{K_{H}} = \sqrt{\frac{1}{9}} \Rightarrow h = 0.25$

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Ionization constants of weak acid HA and weak base BOH are $3.0 \times 10^{- 7}$ each at 298 K. The percent degree of hydrolysis of their salt at the dilution of 10 L is :

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Ionization constants of weak acid HA and weak base BOH are 3.0×10−7 each at 298 K. The percent degree of hydrolysis of their salt at the dilution of 10 L is :

The correct option is D 25 kH=KwKaKb=1×0×10−14(3×10−7)2=19h1−h=√KH=√19⇒h=0.25

Ionization constants of weak acid HA and weak base BOH are $3.0 \times 10^{- 7}$ each at 298 K. The percent degree of hydrolysis of their salt at the dilution of 10 L is :

The correct option is D
25

$k_{H} = \frac{K_{w}}{K_{a} K_{b}} = \frac{1 \times 0 \times 10^{- 14}}{(3 \times 10^{- 7})^{2}} = \frac{1}{9} \frac{h}{1 - h} = \sqrt{K_{H}} = \sqrt{\frac{1}{9}} \Rightarrow h = 0.25$