Question

$NaOH$ is a strong base. What will be the $pH$ of $5.0\times {10}^{-2}MNaOH$ solution?

$(\log 2=0.3)$

• A. $14.00$
• B. $13.70$
• C. $13.00$
• D. $12.70$

Answer 1

Answer: (D)

$12.70$

Given that,
$5.0\times {10}^{-2}MNaOH\equiv \left[O{H}^{-}\right]=5\times {10}^{-2}M$

$\left[H^{+}\right]\left[O{H}^{-}\right]=1\times {10}^{-14}$

$\left[H^{+}\right]\times 5\times {10}^{-2}=1\times {10}^{-14}$

$\left[H^{+}\right]=\frac{1\times {10}^{-14}}{5\times {10}^{-2}}=2\times {10}^{-13}$

$pH=-\log \left[H^{+}\right]=-\log (2\times {10}^{-13})$

$=12.69\approx 12.70$.

Option D is correct.