Question

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

• A. $ 5 $
• B. $ 3 $
• C. $ 9 $
• D. $ 11 $

Answer 1

The Correct Option is C

$\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 $