Question

Given that, (in ${\text{S cm}}^2{\text{eq}}^{-1}$) at $T=298K$ : ${\Lambda }_{eq}^o$ for $Ba(OH{)}_2,BaC{l}_2$ and $N{H}_{4}Cl$ are $228.8,120.3$ and $129.8$ respectively. Specific conductance for $0.2NN{H}_{4}OH$ solution is $4.766\times {10}^{-4}{\text{S cm}}^{-1}$, then the value of $pH$ of the given solution of $N{H}_{4}OH$ will be nearly. (Take $log2=0.3$)

Answer 1

${\Lambda }_{eq}^{o}Ba(OH{)}_2={\lambda }_{eq}^{o}B{a}^{2+}+{\lambda }_{eq}^{o}O{H}^{-}...(i)$
${\Lambda }_{eq}^{o}BaC{l}_2={\lambda }_{eq}^{o}B{a}^{2+}+{\lambda }_{eq}^{o}C{l}^{-}...\left(ii\right)$
${\Lambda }_{eq}^{o}N{H}_{4}Cl={\lambda }_{eq}^{o}N{H}_4^{+}+{\lambda }_{eq}^{o}C{l}^{-}...\left(iii\right)$
${\Lambda }_{eq}^{o}N{H}_{4}OH={\lambda }_{eq}^{o}N{H}_4^{+}+{\lambda }_{eq}^{o}O{H}^{-}...\left(iv\right)$
By equating equations $\left(i\right)+\left(iii\right)-\left(ii\right)$
${\lambda }_{eq}^{o}N{H}_{4}OH=(228.8+129.8)-120.3=238.33{\text{cm}}^2{\text{eq}}^{-1}$
${\lambda }_{eq}N{H}_{4}OH=\frac{4.766\times {10}^{-4}\times 1000}{0.2}=2.383$
$\alpha =\frac{{\lambda }_{eq}}{{\lambda }_{eq}^o}={10}^{-2}$
$$\begin{array}{cccc}N{H}_{4}OH& \rightleftharpoons & N{H}_4^{+}& +O{H}^{-}\\ c(1-\alpha )& & c\alpha & c\alpha \end{array}$$
$\left[O{H}^{-}\right]=0.2\times {10}^{-2}=2\times {10}^{-3}$
$pOH=3-log\left(2\right)pH=14-3-log\left(2\right)pH=11.3$