Question

5.6 grams of KOH $(M.wt=56)$ is present in 1 litre of solution. Its pH is:

• A. 1
• B. 13
• C. 14
• D. 0

Answer 1

Answer: (B)

13

Molarity of KOH solution = $=\frac{\text{Weight}}{\text{Molecular weight}\times \text{volume}}=\frac{5.6}{56\times 1}=0.1$ M.

This is equal to $\left[O{H}^{-}\right]$.

$pOH=-log\left[O{H}^{-}\right]=-log\left(0.1\right)=1$

$pH=14-pOH=14-1=13$.

Hence, option B is correct.