Question

Conductivity of \[{\rm{0}}{\rm{.00241}}\,{\rm{M}}\]acetic acid solution is \[{\rm{7}}{\rm{.896 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{Sc}}{{\rm{m}}^{{\rm{ - 1}}}}\] and \[{\Lambda _0}\]for acetic acid is \[{\rm{390}}{\rm{.5}}\,{\rm{Sc}}{{\rm{m}}^{\rm{2}}}{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\]. The value of molar conductivity and dissociation constant are, respectively:
A. \[12.2\,{\rm{Sc}}{{\rm{m}}^{\rm{2}}}{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}},1.11 \times {10^{ - 4}}\]
B. \[10.7\,{\rm{Sc}}{{\rm{m}}^{\rm{2}}}{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}},2.85 \times {10^{ - 4}}\]
C. \[22.7\,{\rm{Sc}}{{\rm{m}}^{\rm{2}}}{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}},1.85 \times {10^{ - 5}}\]
D. \[32.76\,{\rm{Sc}}{{\rm{m}}^{\rm{2}}}{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}},1.85 \times {10^{ - 5}}\]

Answer 1

Hint: Specific conductance or conductivity is defined as the conductance of one unit cube (\[{\rm{1}}\,{\rm{c}}{{\rm{m}}^{\rm{3}}}\]or \[{\rm{1}}\,{{\rm{m}}^{\rm{3}}}\] )of a conductor. For electrolytes, the specific conductance is defined as the conductance of a solution of electrolyte placed in between two electrodes having an area of \[{\rm{1 c}}{{\rm{m}}^{\rm{2}}}\]and separated by a distance of \[{\rm{1 cm}}\].

Formula used Molar conductivity can be calculated by using the formula as shown below:
$\Lambda_m = K \times V\\
= K \times \dfrac{1000}{c}\\
= K \times \dfrac{1000}{molarity}$
where, \[K\]= specific conductivity
\[{\rm{V}}\]= volume of the solution containing one mole of the electrolyte
\[{\rm{C}}\] = molar concentration
Dissociation constant (\[{{\rm{K}}_{\rm{c}}}\]) can be calculated by using the relationship as shown below:
\[{{\rm{K}}_{\rm{c}}} = \,\dfrac{{c{\alpha ^2}}}{{1 - \alpha }}\]
where, \[c\]= concentration
\[\alpha \]= degree of dissociation

Complete Step by Step Solution:
The molar conductivity of a solution at a dilution \[{\rm{V}}\] is the conductance of all the ions produced from one mole of the electrolyte dissolved in \[{\rm{V}}\,{\rm{c}}{{\rm{m}}^3}\]of the solution when the electrodes are \[{\rm{1 cm}}\] apart and the area of the electrodes is so large that the whole of the solution is contained between them. It is usually expressed by \[{\Lambda _{\rm{m}}}\].

Mathematically,
$\Lambda_m = K \times V\\
= K \times \dfrac{1000}{c}\\
= K \times \dfrac{1000}{molarity}$
where, \[K\]= specific conductivity
\[{\rm{V}}\]= volume of the solution containing one mole of the electrolyte
\[{\rm{C}}\] = molar concentration

Units of molar conductivity (\[{\Lambda _{\rm{m}}}\]) is \[{\rm{oh}}{{\rm{m}}^{{\rm{ - 1}}}}{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\] or \[{\rm{Sc}}{{\rm{m}}^{\rm{2}}}{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\].

The \[{\Lambda ^0}_{\rm{m}}\] values can also be used to calculate the degree of dissociation (\[{\alpha _c}\]) and also the dissociation constants of weak acids and weak bases. According to Arrhenius, we have,
\[{\alpha _c} = \dfrac{{{\Lambda ^c}_{\rm{m}}}}{{{\Lambda ^0}_{\rm{m}}}}\]
where, \[{\Lambda ^0}_{\rm{m}}\]is the molar conductivity of the electrolyte at infinite dilution \[c\]and \[{\alpha _c}\]is its degree of dissociation at concentration \[c\]. By knowing the degree of dissociation (\[\alpha \]), the dissociation constant (\[{{\rm{K}}_{\rm{c}}}\]) of a weak electrolyte can be calculated by using the relation as shown below:
\[{{\rm{K}}_{\rm{c}}} = \,\dfrac{{c{\alpha ^2}}}{{1 - \alpha }}\]

As per given data,
Molarity = \[{\rm{0}}{\rm{.00241}}\,{\rm{M}}\]
Conductivity (\[K\]) = \[{\rm{7}}{\rm{.896 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{Sc}}{{\rm{m}}^{{\rm{ - 1}}}}\]
\[{\Lambda _0}\]= \[{\rm{390}}{\rm{.5}}\,{\rm{Sc}}{{\rm{m}}^{\rm{2}}}{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\]

Now,
$\Lambda_m = K \times V\\
= K \times \dfrac{1000}{c}\\$
${\rm{ = }}\,K \times \dfrac{{1000}}{{{\rm{molarity}}}}$
$= \dfrac{7.896 \times 10^{-5} Scm^{-1} \times 1000}{0.00241 \ M}$
$= 32.7634 Scm^{2}mol^{-1}$

Hence, molar conductivity (\[{\Lambda _{\rm{m}}}\]) of acetic acid is found to be \[32.7634\,{\rm{Sc}}{{\rm{m}}^{\rm{2}}}{\rm{mo}}{{\rm{l}}^{{\rm{ - 1}}}}\].
Again, find the degree of dissociation (\[{\alpha _c}\]) as:
${\alpha _c} = \dfrac{{{\Lambda ^c}_{\rm{m}}}}{{{\Lambda ^0}_{\rm{m}}}}\\$
$ = \dfrac{32.7634 Scm^2mol^{-1}}{390.5 Scm^2mol^{-1}} = 0.0839$

Use the value of degree of dissociation to solve for dissociation constant (\[{{\rm{K}}_{\rm{c}}}\]) as:
$K_c = \dfrac{c\alpha^2}{1-\alpha}\\
= \dfrac{0.00241\times(0.0839)^2}{1-0.0839}\\
= \dfrac{0.00241\times7.03921\times10^{-3}}{0.9161}\\
= \dfrac{0.01696\times10^{-3}}{0.9161}\\
= 0.0185\times10^{-3}\\
= 1.85\times 10^{-5}\\$
Hence, dissociation constant (\[{{\rm{K}}_{\rm{c}}}\]) for acetic acid is found to be \[1.85 \times {10^{ - 5}}\]
Therefore, option D is correct.

Note: Molar conductance of an electrolyte also depends upon the concentration of its solution. Usually, molar conductance increases by increase in dilution (or decrease in concentration) of solution. The reason is that on dilution, the dissociation of electrolyte goes on increasing.