Question

The $K_w$ for water for the equillibrium, $2H_{2}O\rightleftharpoons H_3{O}^{+}O{H}^{-}$ changes from ${10}^{-14}$ at $25{}^{\circ }\text{C}$ to $9.62\times {10}^{-14}$ at $60{}^{\circ }\text{C}$.
Calculate the $pH$ and nature at $60{}^{\circ }\text{C}$

• A. $pH=5.51$, solution is basic
• B. $pH=6.51$, solution is neutral
• C. $pH=5.51$, solution is acidic
• D. $pH=6.51$, solution is basic

Answer 1

The correct option is B $pH=6.51$, solution is neutral

$pH=-log\left[H^{+}\right]$, $K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$,
$9.62\times {10}^{-14}=x^2$
$x=\sqrt{9.62\times {10}^{-14}}$
$\Rightarrow x=3.18\times {10}^7\Rightarrow pH=-log[3.18\times {10}^7]\Rightarrow pH=6.51$

$pH=-log\left[H^{+}\right]$
$pH=-log\frac{1}{\left[H^{+}\right]}$
$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
$-log\left[H^{+}\right]-log\left[O{H}^{-}\right]=-log{K}_w$
$pH+pOH=p{K}_w$
At $25{}^{\circ }\text{C},K_w={10}^{-14}$
At $65{}^{\circ }\text{C}$, solution is neutral.