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# How much $A{{g}^{+}}$ would remain in solution after mixing equal volumes of 0.080 M $AgN{{O}_{3}}$ and 0.08 N $HOCN$.[Given that: ${{K}_{sp}}$ for $AgOCN$ = $2.3\text{ x 1}{{\text{0}}^{-7}}$; ${{K}_{a}}(HOCN)$ = $3.3\text{ x 1}{{\text{0}}^{-4}}$](A) $5\text{ x 1}{{\text{0}}^{-5}}\text{ M}$(B) $5\text{ x 1}{{\text{0}}^{-4}}\text{ M}$(C) $5\text{ x 1}{{\text{0}}^{-3}}\text{ M}$(D) $5\text{ x 1}{{\text{0}}^{-2}}\text{ M}$

Last updated date: 11th Aug 2024
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Hint: An attempt to this question can be made by writing the dissociation of silver nitrate and cyanic acid. Now write the reversible reaction between silver ions and the conjugate base of cyanic acid. With this, you can find a relation between the ${{K}_{sp}}$ and ${{K}_{a}}$ values mentioned in the question. From the relation, you can find out the concentration of silver ions left in the solution unreacted.

Complete step by step solution:
Solubility is defined as the property of a substance (solute) to get dissolved in a solvent in order to form a solution.
The solubility product constant is the equilibrium constant for the dissolution of a solid solute in a solvent to form a solution. It is denoted by the symbol ${{K}_{sp}}$. We can form an idea about the dissolution of an ionic compound by the value of its solubility product.
We will write the dissociation of cyanic acid and the reaction of silver ions with the conjugate base.
$HOCN\to \text{ }{{H}^{+}}+OC{{N}^{-}}$
$A{{g}^{+}}+OC{{N}^{-}}\to \text{ }AgOCN(s)$
Amount of $[A{{g}^{+}}]$ precipitated = $(0.04 - x)$ $M$
Amount of $[OC{{N}^{-}}]$ precipitated = $(0.04 - x)$ $M$
\begin{align} & [\text{HOCN }\!\!]\!\!\text{ = x - }\!\![\!\!\text{ OC}{{\text{N}}^{-}}]=\text{ x M} \\ & \text{ }\!\![\!\!\text{ }{{\text{H}}^{+}}]\text{ = 0}\text{.4 - x M} \\ \end{align}
Then,
$\dfrac{[OC{{N}^{-}}]}{[HOCN]}\text{ = }\dfrac{{{K}_{a}}}{[{{H}^{+}}]}\text{ = }\dfrac{3.3\text{ x 1}{{\text{0}}^{-4}}}{0.04-\text{x}}$
From this we can calculate the $[OC{{N}^{-}}]$ as ,
$[OC{{N}^{-}}]\text{ = }\dfrac{\text{2}\text{.3 x 1}{{\text{0}}^{-4}}\text{ x n}}{0.04-\text{x}}$
$[A{{g}^{+}}][OC{{N}^{-}}]\text{ = }\dfrac{3.3\text{ x 1}{{\text{0}}^{-4}}{{x}^{2}}}{0.04-x}\text{ = }{{\text{K}}_{sp}}$
Hence, $\text{x = }\!\![\!\!\text{ A}{{\text{g}}^{+}}]\text{ = 5 x 1}{{\text{0}}^{-3}}\text{ M}$

Therefore, the correct answer is option (C).

Note: The common ion effect describes the effect of adding a common ion on the ​equilibrium of the new solution. The common ion effect generally decreases the ​solubility of a solute. The equilibrium shifts to the left to relieve off the excess product formed.
When a strong electrolyte is added to a solution of a weak electrolyte, the solubility of the weak electrolyte decreases leading to the formation of precipitate at the bottom of the testing apparatus.