Question

${{K}_{a}}$ of acetic acid will be; if 0.05N solution has equivalent conductance of $7.36\,mho\,c{{m}^{2}}$at ${{25}^{0}}C$ (${{\lambda }^{\infty }}_{C{{H}_{3}}COOH}=390.70$)
A.$1.76\times {{10}^{-5}}mole/litre$
B.$2.99\times {{10}^{-5}}mole/litre$
C.$3.11\times {{10}^{-5}}mole/litre$
D. None of these

Answer 1

The concept of dissociation of weak acids and dissociation constant is to be used in this question. In case of a weak acid $C{{H}_{3}}COOH$ we can assume normality equal to its molarity for easy calculation.

Complete step by step solution:
In order to answer this question, we need to learn about the dissociation of acids. The conductivity of a metallic conductor has a definite value at a given temperature. However, in case of electrolytic solution the conductivity not only depends on temperature but also varies with the concentration of the solution. It is due to the fact that conductivity of a solution is attributed to the presence of ions. The number of ions in one centimetre cube of the solution changes with the variation in the concentration of solution. Therefore, in case of electrolytic solutions, a more appropriate term, molar conductivity is used, which takes into account the concentration of the solution as well as its conductivity.
Molar conductivity is defined as the conductance of a solution kept between the electrodes at unit distance apart and having an area of cross-section large enough to accommodate sufficient volume of the solution that contains one mole of electrolyte. It is denoted by the symbol ${{\lambda }_{m}}$. The degree of dissociation is defined as the ratio of molar conductivity to the molar conductivity at infinite dilution. According to the data given in the question, the degree of dissociation is$\dfrac{7.36}{390.7}=0.0188$. Now, let us see the equilibrium:

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{{H}_{3}}COOH\rightleftharpoons C{{H}_{3}}CO{{O}^{-}}+\,\,\,{{H}^{+}} \\\ & at\,t=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.05\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\\ & at=t'\,\,\,\,\,\,\,\,\,\,\,0.05(1-x)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.05x\,\,\,\,\,\,\,\,\,\,\,0.05x \\\ \end{aligned}$$ So, we can calculate the value of ${{K}_{a}}$as: $${{K}_{a}}=\dfrac{0.05x\,\times \,0.05x}{0.05(1-x)}$$ However, the term (1-x) will be equal to 1 as x is very small. So, ${{K}_{a}}=0.05{{x}^{2}}=0.05\times {{(0.0188)}^{2}}=1.76\times {{10}^{-5}}mole/litre$ **Hence, we get our correct answer as option A.** **NOTE:** Equivalent conductivity of a solution is defined as the conductance of the solution placed between two parallel electrodes one centimetre apart and having an area of cross section large enough to accommodate the entire solution which contains one g-equivalent of electrolyte. It is denoted by ${{\lambda }_{eq}}$