JEE Advance - Chemistry (2017 - Paper 1 Offline - No. 6)
The conductance of a $$0.0015$$ $$M$$ aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized $$Pt$$ electrodes. The distance between the electrodes is $$120$$ $$cm$$ with an area of cross section of $$1$$ $$c{m^2}.$$ The conductance of this solution was found to be $$5 \times {10^{ - 7}}S.$$ The $$pH$$ of the solution is $$4.$$ The value of limiting molar conductivity $$\left( {\Lambda _m^o} \right)$$ of this weak monobasic acid in aqueous solution is $$Z \times {10^2}S$$ $$c{m^2}$$ $$mo{l^{ - 1}}.$$ The value of $$Z$$ is
Answer
6
Explanation
Given:
(i) Concentration of weak monobasic $\operatorname{acid}(\mathrm{C})=0.0015 \mathrm{M}$
(ii) Distance between the electrodes (d) $=120 \mathrm{~cm}$
(iv) Conductance of solution of monobasic acid $(G)=5 \times 10^{-7} \mathrm{~S}$
(v) $\mathrm{pH}$ of the solution $=4$
To Find: The value of $\mathrm{Z}$ in the limiting molar conductivity $\left(\Lambda .{ }^{\circ} \mathrm{m}\right) \mathrm{Z} \times 10^2 \mathrm{~S} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1}$ Formula used:
(i) $$ \mathrm{K}=\frac{\mathrm{G} \times l}{\mathrm{~A}} $$
(ii) $$ \left[\mathrm{H}_3 \mathrm{O}^{+}\right]^{\mathrm{A}}=10^{-\mathrm{pH}} $$
(iii) $$ \alpha=\frac{\Lambda_m^{\mathrm{C}}}{\Lambda_m^0} $$
Calculations: The conductivity of the aqueous solution of weak monobasic acid is represented as:
$$ \begin{aligned} & \mathrm{K}=\frac{\mathrm{G} \times l}{\mathrm{~A}}=\frac{5 \times 10^{-7} \mathrm{~S} \times 120 \mathrm{~cm}}{1 \mathrm{~cm}^2} \\\\ & \mathrm{~K}=6 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1} \end{aligned} $$
The molar conductivity at concentration $(\mathrm{C}=0.0015 \mathrm{M})$ of weak monobasic acid is:
$$ \begin{gathered} \Lambda_m^c=\frac{\mathrm{K} \times 1000}{\mathrm{C}}=\frac{6 \times 10^{-7} \mathrm{~S} \mathrm{~cm}^{-1} \times 1000}{0.0015 \mathrm{M}} \\\\ \Lambda_m^c=40 \mathrm{~S} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1} ~~~~....(i) \end{gathered} $$
Weak monobasic acid dissociated to give monobasic anion and hydronium ion as follows:
$$ \mathrm{HA}(a q) \rightarrow \mathrm{H}_3^{+} \mathrm{O}(a q)+\mathrm{A}^{-}(a q) $$
$\alpha=$ degree of hydrolysis of weak monobasic acid
$$ \begin{aligned} \mathrm{C} \alpha & =\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=10^{-\mathrm{pH}} \\\\ & =10^{-\mathrm{H}} \mathrm{M} \\\\ \alpha & =\frac{10^{-4}}{0.0015 \mathrm{M}} ~~~~....(ii) \end{aligned} $$
We know, degree of hydrolysis is the ratio of molar conductance at concentration $\mathrm{C}$ to molar conductance at infinite dilution.
$$ \mathrm{\alpha}=\frac{\Lambda_m^{\mathrm{C}}}{\Lambda_m^0} $$
Substituting the value of and from equation (i) and (ii) respectively.
$$ \begin{aligned} & \frac{10^{-4}}{0.0015} =\frac{40}{\Lambda_m^0} \mathrm{~S} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1} \\\\ &\Lambda_m^0 =\frac{40 \times 0.0015}{10^{-4}} \mathrm{~S} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1} \\\\ & =6 \times 10^2 \mathrm{~S} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1} \end{aligned} $$
The value of limiting molar conductivity is $6 \times 10^2 \mathrm{~S} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1}$, where the value of $Z=6$
(i) Concentration of weak monobasic $\operatorname{acid}(\mathrm{C})=0.0015 \mathrm{M}$
(ii) Distance between the electrodes (d) $=120 \mathrm{~cm}$
(iv) Conductance of solution of monobasic acid $(G)=5 \times 10^{-7} \mathrm{~S}$
(v) $\mathrm{pH}$ of the solution $=4$
To Find: The value of $\mathrm{Z}$ in the limiting molar conductivity $\left(\Lambda .{ }^{\circ} \mathrm{m}\right) \mathrm{Z} \times 10^2 \mathrm{~S} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1}$ Formula used:
(i) $$ \mathrm{K}=\frac{\mathrm{G} \times l}{\mathrm{~A}} $$
(ii) $$ \left[\mathrm{H}_3 \mathrm{O}^{+}\right]^{\mathrm{A}}=10^{-\mathrm{pH}} $$
(iii) $$ \alpha=\frac{\Lambda_m^{\mathrm{C}}}{\Lambda_m^0} $$
Calculations: The conductivity of the aqueous solution of weak monobasic acid is represented as:
$$ \begin{aligned} & \mathrm{K}=\frac{\mathrm{G} \times l}{\mathrm{~A}}=\frac{5 \times 10^{-7} \mathrm{~S} \times 120 \mathrm{~cm}}{1 \mathrm{~cm}^2} \\\\ & \mathrm{~K}=6 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1} \end{aligned} $$
The molar conductivity at concentration $(\mathrm{C}=0.0015 \mathrm{M})$ of weak monobasic acid is:
$$ \begin{gathered} \Lambda_m^c=\frac{\mathrm{K} \times 1000}{\mathrm{C}}=\frac{6 \times 10^{-7} \mathrm{~S} \mathrm{~cm}^{-1} \times 1000}{0.0015 \mathrm{M}} \\\\ \Lambda_m^c=40 \mathrm{~S} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1} ~~~~....(i) \end{gathered} $$
Weak monobasic acid dissociated to give monobasic anion and hydronium ion as follows:
$$ \mathrm{HA}(a q) \rightarrow \mathrm{H}_3^{+} \mathrm{O}(a q)+\mathrm{A}^{-}(a q) $$
$\alpha=$ degree of hydrolysis of weak monobasic acid
$$ \begin{aligned} \mathrm{C} \alpha & =\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=10^{-\mathrm{pH}} \\\\ & =10^{-\mathrm{H}} \mathrm{M} \\\\ \alpha & =\frac{10^{-4}}{0.0015 \mathrm{M}} ~~~~....(ii) \end{aligned} $$
We know, degree of hydrolysis is the ratio of molar conductance at concentration $\mathrm{C}$ to molar conductance at infinite dilution.
$$ \mathrm{\alpha}=\frac{\Lambda_m^{\mathrm{C}}}{\Lambda_m^0} $$
Substituting the value of and from equation (i) and (ii) respectively.
$$ \begin{aligned} & \frac{10^{-4}}{0.0015} =\frac{40}{\Lambda_m^0} \mathrm{~S} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1} \\\\ &\Lambda_m^0 =\frac{40 \times 0.0015}{10^{-4}} \mathrm{~S} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1} \\\\ & =6 \times 10^2 \mathrm{~S} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1} \end{aligned} $$
The value of limiting molar conductivity is $6 \times 10^2 \mathrm{~S} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1}$, where the value of $Z=6$
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