Question

From the following molar conductivities at infinite dilution,
$\Lambda _{{\text{m }}}^{\text{o}}{\text{for A}}{{\text{l}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}{\text{ = }}858{\text{ S c}}{{\text{m}}^{\text{2}}}{\text{ mo}}{{\text{l}}^{{\text{ - 1}}}}$
$\Lambda _{{\text{m }}}^{\text{o}}{\text{for N}}{{\text{H}}_4}{\text{OH = }}238.3{\text{ S c}}{{\text{m}}^{\text{2}}}{\text{ mo}}{{\text{l}}^{{\text{ - 1}}}}$
$\Lambda _{{\text{m }}}^{\text{o}}{\text{for (N}}{{\text{H}}_{\text{4}}}{{\text{)}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ = }}238.4{\text{ S c}}{{\text{m}}^{\text{2}}}{\text{ mo}}{{\text{l}}^{{\text{ - 1}}}}$
calculate $\Lambda _{{\text{m }}}^{\text{o}}{\text{for Al(OH}}{{\text{)}}_{\text{3}}}$.
A.$715.2{\text{ S c}}{{\text{m}}^{\text{2}}}{\text{ mo}}{{\text{l}}^{{\text{ - 1}}}}$
B.$1575.6{\text{ S c}}{{\text{m}}^{\text{2}}}{\text{ mo}}{{\text{l}}^{{\text{ - 1}}}}$
C.$786.3{\text{ S c}}{{\text{m}}^{\text{2}}}{\text{ mo}}{{\text{l}}^{{\text{ - 1}}}}$
D.$157.56{\text{ S c}}{{\text{m}}^{\text{2}}}{\text{ mo}}{{\text{l}}^{{\text{ - 1}}}}$

Answer 1

To solve this question, it is required to have knowledge about the Kohlraush law. It states that at infinite dilution, the equivalent conductivity of an electrolyte is equal to the sum of the conductance of the cations and anions (formula given). We shall use this law and write the conductivity of the given compounds as the sum of their ions. Then we shall substitute the values possible to calculate the $\Lambda _{{\text{m }}}^{\text{o}}{\text{for Al(OH}}{{\text{)}}_{\text{3}}}$ .
Formula used: $\Lambda _{\text{m}}^{\text{o}}{\text{ = }}{{\text{z}}_{\text{ + }}}\Lambda _{\text{ + }}^{\text{o}}{\text{ + }}{{\text{z}}_{\text{ - }}}\Lambda _{\text{ - }}^{\text{o}}$ where $\Lambda _m^o$ is the conductivity of the electrolyte, $\Lambda _ + ^o$ is the conductivity of the cation, $\Lambda _ - ^o$ is the conductivity of the anion, ${{\text{z}}_{\text{ + }}}$ is the stoichiometric coefficient of the cation and ${{\text{z}}_{\text{ - }}}$ is the stoichiometric coefficient of the anion.

Complete step by step answer:
The Kohlraush law can be used to determine the limiting molar conductivities for an electrolyte. Weak electrolytes have lower molar conductivities and lower degree of dissociation at higher concentration and increase steeply on dilution. Here, Kohlraush law is used to calculate the molar conductivity of such electrolytes.
${\text{A}}{{\text{l}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}$ in a solution dissociates into:
${\text{A}}{{\text{l}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}} \to {\text{2A}}{{\text{l}}^{{\text{3 + }}}}{\text{ + 3SO}}_{\text{4}}^{{\text{2 - }}}$
So, applying Kohlraush law we get:
$\Lambda _{{\text{A}}{{\text{l}}_{\text{2}}}{{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}}_{\text{3}}}{\text{ }}}^{\text{o}}{\text{ = }}2\Lambda _{{\text{A}}{{\text{l}}^{{\text{3 + }}}}{\text{ }}}^{\text{o}}{\text{ + }}3\Lambda _{{\text{SO}}_{\text{4}}^{{\text{2 - }}}{\text{ }}}^{\text{o}}$ (Eq. 1)
${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}}$ in a solution dissociates into:
${\text{N}}{{\text{H}}_{\text{4}}}{\text{OH}} \to {\text{NH}}_{\text{4}}^{\text{ + }}{\text{ + O}}{{\text{H}}^{\text{ - }}}$
So, applying Kohlraush law we get:
$\Lambda _{{\text{ N}}{{\text{H}}_{\text{4}}}{\text{OH }}}^{\text{o}}{\text{ = }}\Lambda _{{\text{NH}}_{\text{4}}^{\text{ + }}{\text{ }}}^{\text{o}}{\text{ + }}\Lambda _{{\text{O}}{{\text{H}}^{\text{ - }}}{\text{ }}}^{\text{o}}$ (Eq. 2)
${{\text{(N}}{{\text{H}}_{\text{4}}}{\text{)}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ in a solution dissociates into:
${{\text{(N}}{{\text{H}}_{\text{4}}}{\text{)}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to {\text{2NH}}_{\text{4}}^{\text{ + }}{\text{ + SO}}_{\text{4}}^{{\text{2 - }}}$
So, applying the Kohlraush law we get:
$\Lambda _{{\text{ (N}}{{\text{H}}_{\text{4}}}{{\text{)}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ }}}^{\text{o}}{\text{ = }}2\Lambda _{{\text{NH}}_{\text{4}}^{\text{ + }}{\text{ }}}^{\text{o}}{\text{ + }}\Lambda _{{\text{SO}}_{\text{4}}^{{\text{2 - }}}{\text{ }}}^{\text{o}}$ (Eq. 3)
${\text{Al(OH}}{{\text{)}}_{\text{3}}}$ in a solution will dissociate into:
${\text{Al(OH}}{{\text{)}}_{\text{3}}} \to {\text{A}}{{\text{l}}^{{\text{3 + }}}}{\text{ + 3O}}{{\text{H}}^{\text{ - }}}$
So, applying Kohlraush law we get:
$\Lambda _{{\text{ Al(OH}}{{\text{)}}_{\text{3}}}{\text{ }}}^{\text{o}}{\text{ = }}\Lambda _{{\text{A}}{{\text{l}}^{{\text{3 + }}}}}^{\text{o}}{\text{ + }}3\Lambda _{{\text{O}}{{\text{H}}^{\text{ - }}}{\text{ }}}^{\text{o}}$ (Eq. 4)
Now, by adding Eq. 1 and six times Eq. 2 and subtracting it from three times Eq. 3, we get:
$2\Lambda _{{\text{A}}{{\text{l}}^{{\text{3 + }}}}{\text{ }}}^{\text{o}}{\text{ + }}3\Lambda _{{\text{SO}}_{\text{4}}^{{\text{2 - }}}{\text{ }}}^{\text{o}}{\text{ + }}6\Lambda _{{\text{NH}}_{\text{4}}^{\text{ + }}{\text{ }}}^{\text{o}}{\text{ + 6\Lambda }}_{{\text{O}}{{\text{H}}^{\text{ - }}}{\text{ }}}^{\text{o}} - {\text{ }}6\Lambda _{{\text{NH}}_{\text{4}}^{\text{ + }}{\text{ }}}^{\text{o}} - {\text{ }}3\Lambda _{{\text{SO}}_{\text{4}}^{{\text{2 - }}}{\text{ }}}^{\text{o}} = 2\Lambda _{{\text{A}}{{\text{l}}^{{\text{3 + }}}}}^{\text{o}}{\text{ + }}6\Lambda _{{\text{O}}{{\text{H}}^{\text{ - }}}{\text{ }}}^{\text{o}}$
$\Rightarrow \Lambda _{{\text{A}}{{\text{l}}_{\text{2}}}{{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}}_{\text{3}}}{\text{ }}}^{\text{o}}{\text{ + }}6\Lambda _{{\text{ N}}{{\text{H}}_{\text{4}}}{\text{OH }}}^{\text{o}} - {\text{ }}3\Lambda _{{\text{ (N}}{{\text{H}}_{\text{4}}}{{\text{)}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ }}}^{\text{o}}{\text{ = }}2\Lambda _{{\text{Al(OH}}{{\text{)}}_{\text{3}}}}^{\text{o}}$
So, by substituting the values in the above equation we get:
$2\Lambda _{{\text{Al(OH}}{{\text{)}}_{\text{3}}}}^{\text{o}}{\text{ = }}858{\text{ }} + {\text{ }}6(238.3){\text{ }} - {\text{ }}3(288.4)$
Solving this, we get:
$2\Lambda _{{\text{Al(OH}}{{\text{)}}_{\text{3}}}}^{\text{o}}{\text{ = }}1572.6$
$\Rightarrow \Lambda _{{\text{Al(OH}}{{\text{)}}_{\text{3}}}}^{\text{o}}{\text{ = }}786.3{\text{ S c}}{{\text{m}}^{\text{2}}}{\text{ mo}}{{\text{l}}^{{\text{ - 1}}}}$

$\therefore$ The correct option is option C, i.e. $786.3{\text{ S c}}{{\text{m}}^{\text{2}}}{\text{ mo}}{{\text{l}}^{{\text{ - 1}}}}$ .

Note:
Kohlraush law is used like algebraic equations, we add, multiply, subtract to get the molar conductivity of the ions required. It can be solved easily if we consider the ions as variables and solve them similar to algebraic equations.