Question

For a saturated solution of AgCl at ${{25}^{\circ }}C$, $K =3.4\times {{10}^{-6}}oh{{m}^{-1}}c{{m}^{-1}}$ and that of ${{H}_{2}}O(l)$  used is $2.02\times {{10}^{-6}}oh{{m}^{-1}}c{{m}^{-1}}$. $\lambda _{m}^{\circ }$  for AgCl is $138oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}$ .The solubility of AgCl in moles per liter is $a\times {{10}^{-5}}$ . Find a.

Answer 1

Hint: Attempt this question by calculating specific conductance for pure AgCl and use in the formula given below. Also we know that for a saturated solution of sparingly soluble salt, molar conductance at any concentration is equal to molar conductance at dilute solution and also solubility of AgCl is equal to its normality.

Formula used: We will use the following formula:-
${{\lambda }_{m}}=\dfrac{K \times 1000}{M}$
$M=\dfrac{N}{n-factor}$

Complete step-by-step answer:
- Specific Conductance (denoted by ‘$K$’- kappa) is defined as the ability of a substance to conduct electricity. It is the reciprocal of specific resistance ($\rho $).
- Molar conductivity is often described as the conductance of all ions produced by one mole of an electrolyte present in a fixed volume of the solution. Mathematically, it can be calculated as follows:-
${{\lambda }_{m}}=\dfrac{K \times 1000}{M}$
where,
${{\lambda }_{m}}$ = molar conductivity of solution
K = specific conductivity
M = molarity of the solution

- We are given the following values:-
Specific conductivity of saturated solution ofAgCl = $K =3.4\times {{10}^{-6}}oh{{m}^{-1}}c{{m}^{-1}}$
Specific conductivity of ${{H}_{2}}O(l)$= $K =2.02\times {{10}^{-6}}oh{{m}^{-1}}c{{m}^{-1}}$
Molar conductivity at infinite dilution ($\lambda _{m}^{\circ }$) = $138oh{{m}^{-1}}c{{m}^{2}}mo{{l}^{-1}}$
- As we know that, for a saturated solution of sparingly soluble salt, molar conductance at any concentration is equal to molar conductance at dilute solution and also solubility of AgCl is equal to its normality. Therefore,
$\lambda _{m}^{\circ }$=${{\lambda }_{m}}$
S (solubility)= Normality
Then, $M=\dfrac{N}{n-factor}=\dfrac{S}{n-factor}$
On replacing these terms in formula of molar conductance, we get:-
$\lambda _{m}^{\circ }=\dfrac{K \times 1000\times n-factor}{S}$
n-factor for AgCl = 1
Therefore, $\lambda _{m}^{\circ }=\dfrac{K \times 1000}{S}$
- Specific conductivity of pure AgCl = specific conductivity of saturated solution of AgCl- specific conductivity of ${{H}_{2}}O(l)$
$K =$ ($3.4\times {{10}^{-6}}oh{{m}^{-1}}c{{m}^{-1}}$ - $2.02\times {{10}^{-6}}oh{{m}^{-1}}c{{m}^{-1}}$)
$K =1.38\times {{10}^{-6}}oh{{m}^{-1}}c{{m}^{-1}}$
- On substituting all the given values in the formula, we get:-
$
 \Rightarrow \lambda _{m}^{{}^\circ }=\dfrac{K \times 1000}{S} \\
 \Rightarrow 138=\dfrac{1.38\times {{10}^{-6}}\times 1000}{S} \\
  \Rightarrow S=\dfrac{138\times {{10}^{-5}}}{138} \\
 \Rightarrow S=1\times {{10}^{-5}} \\
$
- Therefore the solubility of the solution is $S=1\times {{10}^{-5}}$ moles per liter which can be compared to $a\times {{10}^{-5}}$and we get a = 1

Note: -Calculating the solubility of sparingly soluble salt is an application of Kohlrausch’s law.
-Kohlrausch’s law helps us in determining the limiting molar conductivities for any electrolyte (especially weak electrolytes) and also helps us to determine the value of dissociation constant from the value of molar conductivity and limiting molar conductivity for an electrolyte at a given concentration.