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The pH of 0.1 M  NH_3 solution ( K_b = 1.8 \times 10 ^{-5}) is ? 

Option: 1

2.87


Option: 2

11.13


Option: 3

9.13


Option: 4

10.13


Answers (1)

best_answer

Value of Kb -

MOH(aq)\rightleftharpoons M^{+}(aq)+\bar{O}H(aq)

- wherein

K_{b}=\frac{[M^{+}]\:[\bar{O}H]}{[MOH]}


K_{b}=\frac{C\alpha ^{2}}{1-\alpha }

C=initial\:concentration\:of\:base

\alpha =degree\:of\:ionization\:of\:base

 

                      NH_{3}+H_{2}O\rightleftharpoons NH_{4}^{+}+OH^{-}

t=0                 0.1            -                   -               -

t=t                0.1-x                               x               x

 

 

k_b = \frac{x^2}{1-x} \Rightarrow 1.8 \times 10 ^{-5} = \frac{x^2 }{0.1-x}

 

\Rightarrow x^2 = 1.8 \times 10 ^{-6} \Rightarrow x = 1.3 \times 10 ^{-3}M

\\pOH = -log [OH^-] = 2.87 \\ pH = 14-2.87 = 11.13

Posted by

Sanket Gandhi

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