Question

The values of limiting ionic conductance of $H^{+}$ and $HCO{O}^{-}$ ions are respectively $347$ and $58Sc{m}^{2}mo{l}^{-1}$. If the molar conductance of 0.025 M methanoic acid at 298 K is $40Sc{m}^{2}mo{l}^{-1}$, the dissociation constant of methanoic acid at 298 K is:

• A. $1\times {10}^{-5}$
• B. $2\times {10}^{-5}$
• C. $1.5\times {10}^{-4}$
• D. $2.5\times {10}^{-5}$
• E. $2.5\times {10}^{-4}$

Answer 1

Answer: (E)

$2.5\times {10}^{-4}$

Given,

${\wedge }_{H^{+}}^{\circ }=347Sc{m}^{2}mo{l}^{-1}$

${\wedge }_{HCO{O}^{-}}^{\circ }=53Sc{m}^{2}mo{l}^{-1}$

${\wedge }_{HCOOH}^{\circ }=347+53=400Sc{m}^{2}mo{l}^{-1}$

Degree of dissociation,

$\alpha =\frac{{\wedge }_m^{\circ }}{{\wedge }_{HCOOH}^{\circ }}=\frac{40}{400}=\frac{1}{10}=0.1$

$HCOOH$ ionises as:

$\underset{\underset{c-ca}{C}}{HCOOH}\to \underset{\underset{Ca}{0}}{H^{+}}+\underset{\underset{Ca}{0}}{HCO{O}^{-}}$

$K_a=\frac{\left[H^{+}\right]\left[HCO{O}^{-}\right]}{\left[HCOOH\right]}=\frac{C^2{\alpha }^2}{C-C\alpha }=\frac{C{\alpha }^2}{(1-\alpha )}$

$\because \alpha <<1$ so $1-\alpha \approx 1$

$K_a=C{\alpha }^2=0.025\times (0.1{)}^2=2.5\times {10}^{-4}$.