Question

$K_w$ for $H_{2}O$ is $9.62\times {10}^{-14}$ at ${60}^{o}C$. What is $pH$ of water at ${60}^{o}C$.

• A. $3.71$
• B. $6.51$
• C. $7.49$
• D. $11.31$

Answer 1

Answer: (B)

$6.51$

We know that when $K_w={10}^{-14}$,

$\left[H_3{O}^{+}\right]=\left[O{H}^{-}\right]={10}^{-7}mol/d{m}^3$
The value of $K_w$ depends on temperature. So, in the question, the value is not ${10}^{-14}$.
SImilarly, we can use the relation for above given value of $K_w$
So,
$\left[H_3{O}^{+}\right]=\sqrt{9.62\ast {10}^{-14}}=3.1\times {10}^{-7}mol/d{m}^3$
Hence,
$pH=-lo{g}_{10}\left[H_3{O}^{+}\right]=-lo{g}_{10}(3.1\ast {10}^{-7})=7-0.49=6.51$
Shortcut:
Remember that you cannot use log tables in your exam. So, here once you find the value of $\left[H_3{O}^{+}\right]$, taking log will give you "7 - something lesser than 1" which matches with one option only i.e. B.