Question

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

• 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

$pH=-log\left[H^{+}\right]$, $K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$, $9.62\times {10}^{-14}=x^2$
$x=\sqrt{9.62\times {10}^{-14}}$ $\Rightarrow x=3.18\times {10}^7\Rightarrow pH=-log[3.18\times {10}^7]\Rightarrow pH=6.51$
$pHvalue:$ Negative logarithem if $H^{+}$ ion concentration.
$pH=-log\left[H^{+}\right]$
$pH=-log\frac{1}{\left[H^{+}\right]}$
$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
$-log\left[H^{+}\right]-log\left[O{H}^{-}\right]=-log{K}_w$
$pH+pOH=p{K}_w$
at ${25}^{\circ }C,K_w={10}^{-14}$
At ${65}^{\circ }C$ ,solution is neutral.