Question

Given that $\left(inSc{m}^{2}e{q}^{-1}\right)$ at $\text{T}=298\text{K}{\Lambda }_{eq}^{\circ }$ for $Ba(OH{)}_2,BaC{l}_2\text{and}N{H}_{4}Cl$ are 228.8, 120.3 and 129.8 respectively. Specific conductance for 0.2 $N$ $N{H}_{4}OH$ solution is $4.766\times {10}^{-4}{\text{Scm}}^{-1}$, then the $pH$ of given $N{H}_{4}OH$ solution will be:

• A. 9.2
• B. 11.3
• C. 12.1
• D. 7.9

Answer 1

The correct option is B 11.3

${\Lambda }^{\circ }\left(Ba\right(OH{)}_2)={\lambda }_{B{a}^{2+}}^{\circ }+{\lambda }_{O{H}^{-}}^{\circ }......(1)$
${\Lambda }^{\circ }\left(BaC{l}_2\right)={\lambda }_{B{a}^{2+}}^{\circ }+{\lambda }_{C{l}^{-}}^{\circ }......\left(2\right)$
${\Lambda }^{\circ }\left(N{H}_{4}Cl\right)={\lambda }_{N{H}_4^{+}}^{\circ }+{\lambda }_{C{l}^{-}}^{\circ }......\left(3\right)$
${\Lambda }^{\circ }\left(N{H}_{4}OH\right)={\lambda }_{N{H}_4^{+}}^{\circ }+{\lambda }_{O{H}^{-}}^{\circ }......\left(4\right)$
Please note that this only works because we are talking about ${\Lambda }_{eq}^{\circ }$.
$e{q}^n\left(1\right)+e{q}^n\left(3\right)-e{q}^n\left(2\right),$ we get
= $228.8+129.8-120.3=238.3{\text{Scm}}^{2}e{q}^{-1}$
Now, equivalent conductivity of $N{H}_{4}OH$:
${\Lambda }_{eq}=\kappa \times \frac{1000}{\text{Normality}}=\frac{4.766\times {10}^{-4}\times 1000}{0.2}=2.383$
$\alpha =\frac{{\Lambda }_{eq}\left(N{H}_{4}OH\right)}{{\Lambda }^{\circ }\left(N{H}_{4}OH\right)}=\frac{2.383}{238.3}=0.01$ $\stackrel{N{H}_{4}OH}{c(1-\alpha )}\rightleftharpoons \stackrel{N{H}_4^{+}}{c\alpha }+\stackrel{O{H}^{-}}{c\alpha }\left[O{H}^{-}\right]=0.2\times 0.01=2.0\times {10}^{-3}$
$pOH=3-log2=3-0.3010=2.7$
$pH=14-2.7=11.3$