Question

At 25${}^{\circ }C$, ${\lambda }_{(H+)}^0=3.498\times {10}^{-2}$ and , ${\lambda }_{(OH-)}^0=1.98\times {10}^{-2}S{m}^{-2}mo{l}^{-1}$. Given for water $\kappa =5.7\times {10}^{-6}S{m}^{-1}$, Calculate $K_w$.

Answer 1

${\lambda }_{H_{2}O}$ at $25°=\frac{K\times 1000}{C}$

$C=$ Concentration of $H_{2}O$ at $25°C=55.55M$

$\therefore {\lambda }_{H_{2}O}=\frac{5.7\times {10}^{-6}\times 1000}{55.55}=1.02\times {10}^{-4}S{m}^{-2}mo{l}^{-1}$

${\lambda }_{H_{2}O}={\lambda }_{H^{+}}^{\circ }+{\lambda }_{O{H}^{-}}^{\circ }=(3.498+1.98)\times {10}^{-2}=5.478\times {10}^{-2}S{m}^{-2}mo{l}^{-1}$

$\alpha =\frac{\lambda }{\lambda °}=\frac{1.02\times {10}^{-4}}{5.478\times {10}^{-2}}=1.86\times {10}^{-3}$

$K_w=\frac{\left[H^{+}\right]\left[O{H}^{-}\right]}{\left[H_{2}O\right]}=\frac{C\alpha \times C\alpha }{C(1-\alpha )}=\frac{C{\alpha }^2}{1-\alpha }$

$\because \alpha <Cl$

$K_w=C{\alpha }^2=55.55\times (1.86\times {10}^{-3}{)}^2=1.92\times {10}^{-4}$