Question

The equivalent conductivity of a 0.01 N solution of a monobasic acid is 70.0 S $c{m}^{2}e{q}^{-1}$. If the equivalent conductivity of the acid at infinite dilution is 350 $Sc{m}^{2}e{q}^{-1}.$ Then

• A. The degree of dissociation of the acid is 20 %
• B. The dissociation constant of the acid is $5.0\times {10}^{-4}$
• C. The osmotic pressure of the solution is 22.4 cm of Hg at 300K
• D. The pH of the solution is 2.7

Answer 1

Answer: (A), (B), (C), (D)

The pH of the solution is 2.7

$\alpha =\frac{70}{350}=0.2$

$Ka=\frac{C{\alpha }^2}{1-\alpha }=\frac{0.01\times (0.2{)}^2}{1-0.2}=\frac{4\times {10}^{-4}}{0.8}=5\times {10}^{-4}$

$\Pi =iCRT=(1+\alpha )CRT=1.2\times 0.01\times 0.082\times 300\times 76=22.4cmofHg$

$\left[H^{+}\right]=C\alpha =0.01\times 0.2=2\times {10}^{-3}pH=2.7$