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Question: Equal volume of \[0.015M\] \[C{H_3}COOH\] and \[0.015M\] \[NaOH\] are mixed together. What would be ...

Equal volume of 0.015M0.015M CH3COOHC{H_3}COOH and 0.015M0.015M NaOHNaOH are mixed together. What would be the molar conductivity of the mixture if conductivity of CH3COONaC{H_3}COONa is 6.3×10−4Scm−16.3 \times {10^{ - 4}}Sc{m^{ - 1}}.
A. 8.4Scm2mol−18.4Sc{m^2}mo{l^{ - 1}}
B. 84Scm2mol−184Sc{m^2}mo{l^{ - 1}}
C. 4.2Scm2mol−14.2Sc{m^2}mo{l^{ - 1}}
D. 42Scm2mol−142Sc{m^2}mo{l^{ - 1}}

Explanation

Solution

Write down the neutralization reaction that takes place between ethanoic acid and sodium hydroxide to give sodium acetate as the final product. Remember that when equal volumes of two equimolar molar solutions are mixed, the resulting solution has twice the volume that affects its concentration.

Complete answer:
The reaction that takes place between acetic acid and sodium hydroxide is an acid-base reaction in which the neutralization results in the formation of the sodium acetate salt. The chemical equation for the reaction can be written as follows:
CH3COOH+NaOH→CH3COONa+H2OC{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O
Conductance is a phenomenon of carrying charge or allowing electricity to mass through and it requires the mobility of charge species that can act as current carriers. The solutions of electrolytes provide ions and their mobility is responsible for the electrical conductance shown by these solutions. Acids, bases and their corresponding salt solutions are capable of conducting electricity.
The molar conductivity of a solution is defined as the conductivity of an electrolytic solution shown by one of the substances.
The reaction between acetic acid and sodium hydroxide is such that the two react in 1:11:1 molar ratio to produce one mole of the sodium acetate salt. Thus the number of moles remain the same, yet the volume of the solution doubles on mixing.
Thus the molarity of sodium acetate solution can be calculated as follows:
M(CH3COONa)=M1V1(CH3COOH)+M2V2(NaOH)V(CH3COONa)M(C{H_3}COONa) = \dfrac{{{M_1}{V_1}(C{H_3}COOH) + {M_2}{V_2}(NaOH)}}{{V(C{H_3}COONa)}}
M(CH3COONa)=(0.015×V)+(0.015×V)2V=0.0075MM(C{H_3}COONa) = \dfrac{{(0.015 \times V) + (0.015 \times V)}}{{2V}} = 0.0075M
Now, the formula for calculating molar conductivity can be written as follows:
Λm=κ×1000M{\Lambda _m} = \dfrac{{\kappa \times 1000}}{M}
Where, Λm{\Lambda _m} stands for molar conductivity, κ\kappa stands for conductivity of the mixture and MM stands for the molarity of the sodium acetate solution.
Putting in the values we get,
Λm=6.3×10−4Scm−1×10000.0075M=84Scm2mol−1{\Lambda _m} = \dfrac{{6.3 \times {{10}^{ - 4}}Sc{m^{ - 1}} \times 1000}}{{0.0075M}} = 84Sc{m^2}mo{l^{ - 1}}

Hence, the correct option is (B)
Note:
Even though acetic acid is a weak acid nut, sodium hydroxide is a strong base, hence the neutralization reaction between them results in the formation of a strong electrolyte sodium acetate. Due to the formation of a strong electrolyte the reaction goes to completion and no equilibrium is established.