Tag: further aspects of equilibria

Suppose solubility of $AgCNS$ and $AgBr$ in a solution are $a$ and $b$ respectively.

Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.

Assertion: AgCl will not dissolve in a concentrated NaCl solution.
Reason: The chloride ions from NaCl suppress the solubility of AgCl.
This is an example of a common ion effect. The chloride ions are common ions.
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

$NaCl$ is a strong electrolyte. It completely dissociates to form ions. Hence, it is highly soluble in water.
$AgCl$ dissociates to little extent. Hence, it is sparingly soluble in water.
When $NaCl$ is added to a solution of $AgCl$, due to common ion effect (chloride ion is the common ion), the dissociation of $AgCl$ is suppressed. Also as the concentration of chloride ion (from $NaCl$) increases, the ionic product of $AgCl$ exceeds its solubility product and precipitation occurs.

$HA,  \alpha =\sqrt{\displaystyle\frac{10^{-5}}{0.1}}=0.01$ 
(A) On adding $HCl, \left [ HA \right ]=\frac{0.1}{2},\left [ HCl \right ]=\displaystyle\frac{10^{-3}}{2}$
$HCl \rightarrow H^{+}+Cl^{-}$
$HA \rightleftharpoons H^{+}  +A^{-}$
$\displaystyle\frac{0.1}{2}\left ( 1- \alpha  \right )\left (\displaystyle\frac{10^{-3}}{2}+\displaystyle\frac{0.1 \alpha }{2}  \right ) \displaystyle\frac{0.1 \alpha }{2} $
$\Rightarrow 10^{-5}=\displaystyle\frac{\displaystyle\frac{0.1\alpha }{2} \times \left (\displaystyle\frac{10^{-3}}{2}+\displaystyle\frac{0.1 \alpha }{2}  \right ) }{\displaystyle\frac{0.1}{2}\left ( 1-\alpha  \right )}$
Neglecting $\alpha $
$10^{-5}=\alpha \times \left ( \displaystyle\frac{10^{-3}}{2}+\displaystyle\frac{0.1\alpha  }{2} \right )$
$\Rightarrow 0.1 \alpha ^{2}+10^{-3} \alpha -2\times 10^{-5}=0$
$\Rightarrow  \alpha ^{2}+10^{-2}\alpha -2\times 10^{-4}=0$
$\Rightarrow \alpha =\displaystyle\frac{-10^{-2}+\sqrt{10^{-4}+8\times 10^{-4}}}{2}=\displaystyle\frac{2\times 10^{-2}}
{2}=2$
Which is similiar to initial.
(B) In case of two weak acids
$\left [\mathrm H^{+}  \right ]=\sqrt{10^{-5}\times \frac{0.1}{2}+2\times 10^{-6}\times \displaystyle\frac{0.5}{2}}$
$=\sqrt{\displaystyle\frac{10^{-6}}{2}+\displaystyle\frac{10^{-6}}{2}}=10^{-3}$
$\alpha =\displaystyle\frac{10^{-5}}{10^{-3}}=10^{-2}$
which is same as earlier
(C) $HNO _{3}\rightarrow H^{+}+NO _{3}^{-}$
$\displaystyle\frac{0.1}{2}       \displaystyle\frac{0.1M}{2}$
$HA \rightleftharpoons H^{+}   +  A^{-}$
$\displaystyle\frac{0.1M}{2}\left ( 1-\alpha  \right )  \displaystyle\frac{0.1}{2}+\displaystyle\frac{0.1}{2}\alpha                
\displaystyle\frac{0.1}{2}\alpha$
$\Rightarrow 10^{-5}=\displaystyle\frac{\frac{0.1\alpha }{2}\times \displaystyle\frac{0.1 }{2} \left (1+\alpha  \right )  }{\displaystyle\frac{0.1}{2}\left (1-\alpha  \right )}$
Neglecting $\alpha $ due to common ion effect.
$\Rightarrow 10^{-5}=\displaystyle\frac{0.1 \alpha }{2}$
$\Rightarrow \alpha =2\times 10^{-4}$
Hence, it decreases from $0.01  to  2\times 10^{-4}$.

The common ion effect describes the changes that occur with the introduction of ions to a solution containing that same ion.
The role that the common ion effect in solutions is mostly visible in the decrease of solubility of solids. Through the addition of common ions, the solubility of a compound generally decreases due to a shift in equilibrium.
$BaCl _2\rightarrow Ba^{2+} + 2Cl^{-}$
$HCl \rightarrow H^{+} + Cl^{-}$
As $Cl^{-} $ is the common ion so, the $ Ba^{2+}$and $ 2Cl^{-}$ combines ( associate ) to give the undissociated $BaCl _2$. 
Hence , there is increase in concentration of undissociated $BaCl _2$.

$Ag^+ + Cl^- \rightarrow AgCl$
$NaCl \rightarrow Na^+ + Cl^-$
As $Cl^- $ is the common ion in both the solution ( i.e. $NaCl $ and $AgCl$ ) , hence $AgCl$ being weaker electrolyte is precipitated or solubility of the silver chloride would decrease.
Common ion Effect : The common ion effect is responsible for the reduction in the solubility of an ionic precipitate when a soluble compound containing one of the ions of the precipitate is added to the solution in equilibrium with the precipitate. It states that if the concentration of any one of the ions is increased, then, according to Le Chatelier's principle, some of the ions in excess should be removed from solution, by combining with the oppositely charged ions. Some of the salt will be precipitated until the ion product is equal to the solubility product. In short, the common ion effect is the suppression of the degree of dissociation of a weak electrolyte containing a common ion

The correct answer is $(A)$.