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Question: The ionization constant of a weak acid is \(1.6 \times {10^{ - 5}}\) and the molar conductivity at i...

The ionization constant of a weak acid is 1.6×10−51.6 \times {10^{ - 5}} and the molar conductivity at infinite dilution is 380×10−4 S m2 mol−1380 \times {10^{ - 4}}{\text{ S }}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}. If the cell constant is 0.01 m−10.01{\text{ }}{{\text{m}}^{ - 1}} then conductance of 0.01 M0.01{\text{ M}} acid solution is:
(A) 1.52×10−5 S1.52 \times {10^{ - 5}}{\text{ S}}
(B) 1.52×10−7 S1.52 \times {10^{ - 7}}{\text{ S}}
(C) 1.52×10−3 S1.52 \times {10^{ - 3}}{\text{ S}}
(D) 1.52×10−4 S1.52 \times {10^{ - 4}}{\text{ S}}

Explanation

Solution

To solve this we must now the formula for the ionization constant of weak acid in terms of its degree of dissociation. From the formula we can calculate the conductance of the acid.The molar conductivity of both strong and weak electrolytes increases as the dilution increases. This is because dilution increases the degree of dissociation and the total number of ions that carry current increases.

Complete step by step solution: We are given a weak acid. The ionization constant of a weak acid is the product of cell constant and the degree of dissociation of weak acid. Thus,
Ka=Cα2{K_a} = C{\alpha ^2} …… (1)
Where, Ka{K_a} is ionization constant of weak acid,
CC is the concentration of the weak acid,
α\alpha is the degree of dissociation of weak acid.
But we know that the degree of dissociation of a weak electrolyte is the ratio of its molar conductivity to its molar conductivity at infinite dilution. Thus,
α=ΛΛ∘\alpha = \dfrac{\Lambda }{{{\Lambda _ \circ }}}
Where α\alpha is the degree of dissociation,
Λ\Lambda is the molar conductivity of the solution,
Λ∘{\Lambda _ \circ } is the molar conductivity at infinite dilution.
Thus, equation (1) becomes as follows:
Ka=C(ΛΛ∘)2{K_a} = C{\left( {\dfrac{\Lambda }{{{\Lambda _ \circ }}}} \right)^2}
Substitute 1.6×10−51.6 \times {10^{ - 5}} for the ionization constant of weak acid, 0.01 M0.01{\text{ M}} for the concentration of the weak acid, 380×10−4 S m2 mol−1380 \times {10^{ - 4}}{\text{ S }}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}} for the molar conductivity at infinite dilution and solve for the conductance. Thus,
1.6×10−5=0.01 M(Λ380×10−4 S m2 mol−1)21.6 \times {10^{ - 5}} = 0.01{\text{ M}}{\left( {\dfrac{\Lambda }{{380 \times {{10}^{ - 4}}{\text{ S }}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}}}} \right)^2}
⇒(Λ380×10−4 S m2 mol−1)2=1.6×10−50.01 M\Rightarrow {\left( {\dfrac{\Lambda }{{380 \times {{10}^{ - 4}}{\text{ S }}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}}}} \right)^2} = \dfrac{{1.6 \times {{10}^{ - 5}}}}{{0.01{\text{ M}}}}
⇒Λ380×10−4 S m2 mol−1=1.6×10−50.01 M\Rightarrow \dfrac{\Lambda }{{380 \times {{10}^{ - 4}}{\text{ S }}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}}} = \sqrt {\dfrac{{1.6 \times {{10}^{ - 5}}}}{{0.01{\text{ M}}}}}
⇒Λ380×10−4 S m2 mol−1=0.0016\Rightarrow \dfrac{\Lambda }{{380 \times {{10}^{ - 4}}{\text{ S }}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}}} = \sqrt {0.0016}
⇒Λ380×10−4 S m2 mol−1=0.04\Rightarrow \dfrac{\Lambda }{{380 \times {{10}^{ - 4}}{\text{ S }}{{\text{m}}^2}{\text{ mo}}{{\text{l}}^{ - 1}}}} = 0.04
⇒Λ=0.00152×10−3 S\Rightarrow \Lambda = 0.00152 \times {10^{ - 3}}{\text{ S}}
Thus, the conductance of 0.01 M0.01{\text{ M}} acid solution is 1.52×10−3 S1.52 \times {10^{ - 3}}{\text{ S}}.

Thus, the correct option is (C) 1.52×10−3 S1.52 \times {10^{ - 3}}{\text{ S}}.

Note: The conducting power of all the ions that are produced by dissolving one mole of any electrolyte in the solution is known as the molar conductivity of the solution.
As the temperature of the solution increases, the molar conductivity of the solution increases. This is because as the temperature increases the interaction and the mobility of the ions in the solution increases.