Question

Ionic product of water at ${80}^{\circ }C$ is $16\times {10}^{-14}{M}^2$. Find
(a) $PH$ of pure water (Neutral water)
(b) Define a solution acidic/ basic having $PH=6.5\to basic$.

Answer 1

Given, $K_W=16\times {10}^{-14}{M}^2$

Hence $C=\sqrt{16\times {10}^{-4}}{M}^2=4\times {10}^{-7}M$

$pH=-{log}_{10}C=-log(4\times {10}^{-7})=7-log4$

$pH=7-0.6$

$pH=6.4$

$A$. Solution $pH>7$ is basic and $pH<7$ is acidic

Hence if $pH$ is $6.5$, it is acidic.