Question

The $K_w$ for pure water
changes from $1\times {10}^{-14}$ at ${25}^{\circ }C$ to $5.5\times {10}^{-14}$ at ${50}^{\circ }C$. Calculate the $pH$ and nature of water respectively at ${50}^{\circ }C$.
(Take $log\left(2.34\right)=0.37$)

• A. 8.63 and neutral
• B. 8.63 and basic
• C. 6.63 and acidic
• D. 6.63 and neutral

Answer 1

The correct option is D 6.63 and neutral

We know,
$pH=-log\left[H^{+}\right]$,
$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$,
$H_{2}O\left(l\right)\rightleftharpoons H^{+}(aq.)+O{H}^{-}(aq.)$
At equilibrium, in pure water, concentration of $\left[H^{+}\right]=\left[O{H}^{-}\right]$

$\therefore 5.5\times {10}^{-14}=[H^{+}{]}^2$

$\left[H^{+}\right]=\sqrt{5.5\times {10}^{-14}}$

$\Rightarrow \left[H^{+}\right]=2.34\times {10}^{-7}pH=-log(2.34\times {10}^{-7})pH=-(0.37-7)=6.63$
Since, $\left[H^{+}\right]=\left[O{H}^{-}\right]at{50}^{\circ }C$ also
So, nature of the water will remain neutral.



Theory:
As the temperature increases, pH scale also changes.
Neutral solution :
If temperature and solvent are changed then the pH range of the scale will also change.
For example:- pH range is from $0to14$ at ${25}^{o}C$ and $K_w={10}^{-14}$ and neutral pH is 7.
But pH range is from $0to13$ at ${80}^{o}C$ and $K_w={10}^{-13}$ (as with increase in temperature $K_w$ increases) and neutral pH is 6.5.
Note :- pH can also be negative or greater than 14.