Question

Given that (in ohm ${}^{−1}c{m}^{2}e{q}^{−1}$ ); $T=298K$
${λ}_E^{∞}$ for $Ba(OH{)}_2=228.8$ specific conductance
${λ}_E^{∞}$ for $BaC{l}_2=120.3$ for $0.2NN{H}_{4}OH$ solution
${λ}_E^{∞}$ for $N{H}_{4}Cl=129.8$ is $4.766×10^{−4}oh{m}^{−1}c{m}^{−1}$
then value of $pH$ of the given solution of $N{H}_{4}OH$ will be nearly

Answer 1

Answer: 11.3

${λ}_E^{∞}\text{ for }N{H}_{4}OH=228.8−120.3+129.8=238.3$
Now, ${λ}_E=\frac{1000×k}{N}$
$=\frac{1000×4.766×10^{−4}}{0.2}=2.383$
$∴Î\pm =\frac{{λ}_E}{{λ}_E^{∞}}=\frac{2.383}{238.3}=0.01$
$∴O{H}^{−}=Î\pm ×C=.01×0.2=2×10^{−3}$
$∴pOH=3−\log 2=3−0.3=2.7$
$∴pH=14−2.7=11.3$