Following data is given at  T = 298 K 

${\lambda }^{\circ }\text{ eq for }\mathrm{Ba}(\mathrm{OH}{)}_2=228.8{\mathrm{ohm}}^{-1}{\mathrm{cm}}^2{\mathrm{eq}}^{-1}$

${\lambda }^{\circ }\text{ eq for }\mathrm{BaC}{\ell }_2=120.3{\mathrm{ohm}}^{-1}{\mathrm{cm}}^2{\mathrm{eq}}^{-1}$

${\lambda }^0\text{ eq for }{\mathrm{NH}}_4\mathrm{C}\ell =129.8{\mathrm{ohm}}^{-1}{\mathrm{cm}}^2{\mathrm{eq}}^{-1}$

Specific conductance for $0.2{\text{N\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}NH}}_{\text{4}}\text{OH}$ solution is$\text{4.766}\times {\text{10}}^{\text{–4}}{\text{ohm}}^{\text{–1}}{\text{cm}}^{\text{–1}}$

Then the value of pH of the given solution of ${\text{NH}}_{\text{4}}\text{OH}$ will be nearly.