Question

$50$ liters of $0.1MHCl$ is thoroughly mixed with $50$ liters of $0.2MNaOH$. The $pOH$ of the resulting solution is:

• A. 1
• B. 0
• C. 7
• D. 1.3

Answer 1

Answer: (D)

1.3

$NaOH$ is in excess. Hence, it will remain in the solution.

$\left[O{H}^{-}\right]=\frac{molesofNaOH-molesofHCl}{totalvolume}$

Substitute values in the above expression.

$\left[O{H}^{-}\right]=\frac{50\times 0.2-50\times 0.1}{50+50}=0.05$

$pOH=-log\left[O{H}^{-}\right]=-log0.05=1.3$.

Option D is correct.