Question

The specific conductance of a $0.01M$ solution of acetic acid at $298K$ is $1.65 \times {10^{ - 4}}oh{m^{ - 1}}c{m^{ - 1}}$. The molar conductance at infinite dilution for ${H^ + }$ ion and $C{H_3}CO{O^ - }$ ion is $349.1oh{m^{ - 1}}{\text{ and 40}}{\text{.9oh}}{{\text{m}}^{ - 1}}c{m^2}mo{l^{ - 1}}$ respectively. Calculate:
$(i)$Molar conductance of the solution.
$(ii)$ Degree of dissociation of $C{H_3}COOH$
$(iii)$ Dissociation constant for acetic acid

Answer 1

In this question, molar conductance is related with specific conductance that is represented as $K$ and the concentration of the solution. Molar conductance can be given by the following formula:
$\wedge = \dfrac{{1000K}}{C}$

Complete answer:
$(i)$ In this part, when one mole of an electrolyte is dissolved in a solution, the ions produced in the solution are capable of conducting electricity. Therefore, the molar conductance is the conductance of all ions dissolved in one mole of an electrolytic solution. Molar conductance is represented as ($\wedge$). Numerically, the molar conductance is defined as specific conductivity divided by the molar concentration of the solution. It can be given as:
$\wedge = \dfrac{{1000K}}{C}$
Here, $K =$ specific conductivity that is property of the solution which allows it to conduct electricity. It is represented as the inverse of the resistance. It is given as $1.65 \times {10^{ - 4}}oh{m^{ - 1}}c{m^{ - 1}}$ in the question.
$C =$ conductance of the solution that is given as $0.01M$ in the solution.
Now by putting values in the above equation we get,
$\wedge = \dfrac{{1000 \times 1.65 \times {{10}^{ - 4}}}}{{0.01}} \\\ \wedge = 16.5oh{m^{ - 1}}c{m^2}mo{l^{ - 1}} \\\$
$(ii)$In this part, the degree of dissociation($\alpha$) of a solution is the ratio molar conductance ($\wedge$) of the solution to the molar concentration of electrolyte solution at infinite dilution (${ \wedge ^\infty }$). Here the molar concentration at infinite dilution can be given as:
${ \wedge ^\infty } = { \wedge ^\infty }[{H^ + }] + { \wedge ^\infty }[C{H_3}COOH] \\\ = 349.1 + 40.9 \\\ = 390oh{m^{ - 1}}c{m^2}mo{l^{ - 1}} \\\$
Now, the degree of dissociation can be given as:
$\alpha = \dfrac{ \wedge }{{{ \wedge ^\infty }}}$
Where, ${ \wedge ^\infty }$ is the molar concentration at any concentration
${ \wedge ^\infty }$ is the molar concentration at infinite dilution
Now, by putting values in above formula we get,
$\alpha = \dfrac{{16.5}}{{390}} = 0.0423$
$(iii)$ In this part, Dissociation constant (${K_d}$) is the division of the concentration of products to the concentration of the reactants. Now, we have to write the equation of dissociation of acetic acid ($C{H_3}COOH$) that can be given as:
${\text{ }}C{H_3}COOH \rightleftharpoons C{H_3}COO + {H^ + } \\\ Initial{\text{ conc}}{\text{. }}c{\text{ 0 0}} \\\ {\text{Final conc}}{\text{. c(1 - }}\alpha {\text{) c}}\alpha {\text{ c}}\alpha \\\$
$K = \dfrac{{[C{H_3}CO{O^ - }][{H^ + }]}}{{[C{H_3}COOH]}} \\\ = \dfrac{{c\alpha \times c\alpha }}{{c(1 - \alpha )}} \\\ = \dfrac{{0.01 \times {{(0.0423)}^2}}}{{1 - 0.0423}} \\\ = 1.86 \times {10^{ - 5}}mol{L^{ - 1}} \\\$

Note:
Always remember that the Dissociation constant (${K_d}$) has the units of concentration that is $mol{L^{ - 1}}$. The units of molar conductance($\wedge$) is $oh{m^{ - 1}}c{m^2}mo{l^{ - 1}}$ and the degree of dissociation($\alpha$) is also a unitless quantity.