Question

Given the following molar conductivities at ${25}^{\circ }C;HCl,428{\Omega }^{-1}c{m}^{2}mo{l}^{-1};NaCl,$ $128{\Omega }^{-1}c{m}^{2}mo{l}^{-1};NaC\left(sodiumcrotonate\right)$, $85{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$. If the conductivity of a $0.001M$ crotonic acid solution is $3.85\times {10}^{-5}{\Omega }^{-1}c{m}^{-1}$? Then the degree of hydrolysis of ${10}^{-4}M$ solution of sodium crotonate is:

• A. $4.162\times {10}^{-3}$
• B. $3.162\times {10}^{-3}$
• C. $1.162\times {10}^{-3}$
• D. $1.162\times {10}^{-2}$

Answer 1

Answer: (B)

$3.162\times {10}^{-3}$

${\Lambda }_{crotonicacid}^o={\Lambda }_{NaC}^o+{\Lambda }_{HCl}^o-{\Lambda }_{NaCl}^o$

${\Lambda }_{crotonicacid}^o=85+428-128$

$=385$

$K=3.85\times {10}^{-5}Sc{m}^{-1}$

${\Lambda }_m=\frac{3.85\times {10}^{-5}\times 1000}{0.001}$

${\Lambda }_m=38.5$

$\propto =\frac{\Lambda m}{{\Lambda }^{o}m}=\frac{38.5}{385}=0.1$

(A) $Ka=C{\propto }^2=0.001\times (0.1{)}^2$

$K_a={10}^{-3}\times {10}^{-2}={10}^{-5}$

(C) $CH\rightleftharpoons C^{⊝}+H^{\oplus }$

$C^{⊝}+H_{2}O\rightleftharpoons CH+O{H}^{⊝}$

$K_h=\frac{K_w}{K_a}=\frac{{10}^{-14}}{{10}^{-5}}={10}^{-9}$

$h=\sqrt{\frac{K_h}{C}}=\sqrt{\frac{{10}^{-9}}{{10}^{-4}}}=\sqrt{{10}^{-5}}$

$h=\sqrt{10\times {10}^{-6}}=3.162\times {10}^{-3}$