Question

$K_w$ for $H_{2}O$ is $9\times {10}^{-14}$ at ${60}^{o}C$. What is $pH$ of water at ${60}^{o}C$?

$(log3=0.47)$

Answer 1

$K_w=1\times {10}^{-14}$ at ${25}^{0}C$

$K_w=\left[H^{+}\right]\left[{OH}^{-}\right]$, But $\left[H^{+}\right]=\left[{OH}^{-}\right]$ then

$K_w={\left[H^{+}\right]}^2$

Given $K_w=9\times {10}^{-14}$

$9\times {10}^{-14}={\left[H^{+}\right]}^2$

$\Rightarrow \left[H^{+}\right]=3\times {10}^{-7}$

$\Rightarrow pH=-log\left[H^{+}\right]$

$=-log[3\times {10}^{-7}]$

$=7-0.48$

$pH=6.52$