Q. Calculate $pOH\ of\ 0.001\ M\ {NH}_{4}OH$, when it is 1 % dissociated in the solution
A) 5
B) 2.96
C) 9.04
D) 11.4
Explanation
(a)
$\left\lbrack OH^{-} \right\rbrack in\ NH_{4}OH\ solution = C\alpha$
$= 0.001 \times \frac{1}{100} = 1 \times 10^{- 5}$
$pOH = - \log{\lbrack OH^{-}\rbrack}$
$pOH = - \log\left\lbrack 1 \times 10^{- 5} \right\rbrack$
$pOH = 5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.