Question

$K_w$ for $H_{2}O$ is $9.62\times {10}^{-14}$ at ${60}^{o}C$. Calculate the $pH$ of water. Also, what is nature of the solution whose $\left(a\right)pH=6.7$ and $\left(b\right)pH=6.35?$

Answer 1

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

so $[H+][OH-]$ $

$[H+{]}^2=9.62\times {10}^{-14}$

$\left[H^{+}\right]=\sqrt{9.6\times {10}^{-14}}$

$\left[H^{+}\right]=3.06\times {10}^{-7}mold{m}^{-3}$

$pH=-log10\left[H^{+}\right]$

So,

$log3.09\times {10}^{-7}=6.5$

$pH=6.5$ for 60 degrees

(a) for pH = 6.7 water (at ${60}^0$) behaves as acid because its pH is low

(b) for pH = 6.35 water (at ${60}^0$) behaves as base bacause its pH is high