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  4. The degree of dissociation of a 0.01 M weak acid is 10-3 . Its pOH is

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Q. The degree of dissociation of a $ 0.01\, M $ weak acid is $ 10^{-3} $ . Its $ pOH $ is

MHT CETMHT CET 2009

Solution:

$\left[ H ^{+}\right] =\sqrt{K_{a} \cdot C} $

or $=\sqrt{\alpha^{2} \cdot C^{2}}$

$\left[\therefore K_{a}=\alpha^{2} C\right] $

$=\sqrt{\left(10^{-3}\right)^{2} \cdot(0.01)^{2}}$

$=10^{-5}$

$pH =-\log \left[ H ^{+}\right] =-\log 10^{-5} $

$=5$

$pOH =14-5= 9 $