Question

At ${25}^{o}C{\lambda }_{\infty }\left(H^{+}\right)=3.4982\times {10}^{-2}S{m}^2{}^{-1}{\lambda }_{\infty }\left({OH}^{-}\right)=1.98\times {10}^{-2}S{m}^2{}^{-1}$.
Given: Sp. conductance $=5.1\times {10}^{-6}S{m}^{-1}$ for $H_{2}O$, determine $pH$ and $K_w$.

Answer 1

$K_{H_{2}O}=K_{H^{+}}+K_{O{H}^{-}}$ $K=Conductivity$

$K_{{Sul}^n}\times 1000=\left\{{\lambda }_m^{\infty }\right(H^{+})+{\lambda }_m^{\infty }({OH}^{-}\left)\right\}\left[H^{+}\right]$ $\because \underset{Purewater}{\underset{}{[H^{+}=\left[{OH}^{-}\right]]}}$

$\therefore \sqrt{K_W}=\left[H^{+}\right]=\left\{\frac{K_{{Sul}^n}\times 1000}{{\lambda }_m^{\infty }\left(H^{+}\right)+{\lambda }_m^{\infty }\left({OH}^{-}\right)}\right\}=\left\{\frac{5.1\times {10}^{-6}\times 1000\times {10}^{-2}}{(3.4982+1.98){10}^{-2}}\right\}={10}^{-7}$

$pH=-{\log }_{10}\left[H^{+}\right]=7$, $K_W={10}^{-14}=[H^{+}{]}^2$