Question

The $K_w$ for $2H_{2}O\rightleftharpoons H_3{O}^{+}+O{H}^{-}$ changes from ${10}^{-14}$ at ${25}^{o}C$ to $9.62\times {10}^{-14}$ at ${60}^{o}C$. The $pH$ and nature of water at ${60}^{o}C$ will be:
(given: $\sqrt{9.60}=3.10,\log 3.10=0.49$)

• A. $6.51$, neutral
• B. $6.51$, acidic
• C. $6.51$, alkaline
• D. $7.51$, alkaline

Answer 1

The correct option is A $6.51$, neutral

$K_w$ for $H_{2}O$ at ${25}^{0}C={10}^{-14}$
$\left[H^{+}\right]\left[H{O}^{-}\right]={10}^{-14}$ $(K_w=[H^{+}\left]\right[H{O}^{-}]={10}^{-14})$
$\therefore \left[H^{+}\right]={10}^{-7}\text{M}$
$pH=-log\left[H^{+}\right]=-log{10}^{-7}=7$
Now,
$K_w$ for $H_{2}O$ at ${60}^{0}C=9.62\times {10}^{-14}$
For pure water $\left[H^{+}\right]=\left[O{H}^{+}\right]$
$[H^{+}{]}^2$=$9.62\times {10}^{-14}$
$[H+]=\sqrt{9.62\times {10}^{-14}}=3.10\times {10}^{-7}M$
$pH=-\log \left[H^{+}\right]=-\log (3.10\times {10}^{-7})$
$pH=6.51$
Thus, $pH$ of water becomes $6.51$ at ${60}^{o}C$ but the nature is netural since calculation for pure water has been made .