SATHEE: Equilibrium - Result Question 62

Equilibrium - Result Question 62

####64. At $25^{\circ} C$ , the dissociation constant of a base, $B O H$ , is $1.0 \times 10^{- 12}$ . The concentration of hydroxyl ions in $0.01 M$ aqueous solution of the base would be

[2005]

(a) $1.0 \times 10^{- 5} m o l L^{- 1}$

(b) $1.0 \times 10^{- 6} m o l L^{- 1}$

(d) $1.0 \times 10^{- 7} m o l L^{- 1}$

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Solution:

(d) Given $K_{b} = 1.0 \times 10^{- 12}$

$[B O H] = 0.01 M$

$α = \sqrt{K_{b} / c}$

$= \sqrt{\frac{1 \times 10^{- 12}}{0.01}} = 1.0 \times 10^{- 5}$

Now $[O H^{-}] = c . α = 0.01 \times 10^{- 5}$

$= 1 \times 10^{- 7} m o l L^{- 1}$

$K_{a} = \frac{[C H_{3} C O O^{-}] [H^{+}]}{[C H_{3} C O O H]}$

Given that,

$[C H_{3} C O O^{-}] = [H^{+}] = 3.4 \times 10^{- 4} M$

$K_{a}$ for $C H_{3} C O O H = 1.7 \times 10^{- 5}$

$C H_{3} C O O H$ is weak acid, so in it $[C H_{3} C O O H]$ is equal to initial concentration. Hence

$\begin{aligned} 1.7 \times 10^{- 5} = \frac{(3.4 \times 10^{- 4}) (3.4 \times 10^{- 4})}{[C H_{3} C O O H]} \\ \begin{aligned} [C H_{3} C O O H] & = \frac{3.4 \times 10^{- 4} \times 3.4 \times 10^{- 4}}{1.7 \times 10^{- 5}} \\ = 6.8 \times 10^{- 3} M \end{aligned} \end{aligned}$

Equilibrium - Result Question 62

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Equilibrium - Result Question 62

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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