Question

For the reaction; ${\left[ {{\text{Ag}}{{\left( {{\text{CN}}} \right)}_{\text{2}}}} \right]^ - } \to {\text{A}}{{\text{g}}^ + }{\text{ + 2C}}{{\text{N}}^ - }$
The equilibrium constant for the above reaction at $25^\circ {\text{C}}$ is $4.0{\text{ }} \times {10^{ - 19}}$ . Calculate the silver ion concentration in a solution which was originally ${\text{0}}{\text{.10 molar}}$ in ${\text{KCN}}$ and 0.03 molar in {\text{0}}{\text{.10 molar}}$$$${\text{AgN}}{{\text{O}}_3}.
A $7.50{\text{ }} \times {10^{18}}M$
B $7.50{\text{ }} \times {\text{ }}{10^{ - 18}}M$
C $7.50{\text{ }} \times {\text{ }}{10^{10}}M$
D $7.50{\text{ }} \times {\text{ }}{10^{ - 10}}M$

Answer 1

First determine the limiting reagent and obtain the number of moles of the complex ion ${\left[ {{\text{Ag}}{{\left( {{\text{CN}}} \right)}_{\text{2}}}} \right]^ - }$ formed if the limiting reagent completely reacts. Then write an expression for the equilibrium constant, and substitute values in the expression and then calculate the silver ion concentration.

Complete Step by step answer: Initially, the mixture is in ${\text{KCN}}$ and 0.03 molar in ${\text{AgN}}{{\text{O}}_3}$ .
Let the total volume is 1 L, so the molar concentration is equal to the number of moles.
So initially $0.03{\text{ moles}}$ of silver nitrate and $0.10{\text{ moles}}$ of potassium cyanide are present. The number of moles of the complex ion ${\left[ {{\text{Ag}}{{\left( {{\text{CN}}} \right)}_{\text{2}}}} \right]^ - }$ present initially is zero.
Potassium cyanide is present in excess and silver nitrate is the limiting reagent. Assume that silver nitrate completely reacts with potassium cyanide to form the complex ion ${\left[ {{\text{Ag}}{{\left( {{\text{CN}}} \right)}_{\text{2}}}} \right]^ - }$ .
$0.03{\text{ moles}}$ of silver nitrate and $2 \times 0.03{\text{ moles = }}0.06{\text{ moles}}$ of potassium cyanide will form $0.03{\text{ moles}}$ of the complex ion ${\left[ {{\text{Ag}}{{\left( {{\text{CN}}} \right)}_{\text{2}}}} \right]^ - }$ .
$0.10{\text{ moles}} - 0.06{\text{ moles}} = 0.04{\text{ moles}}$ of potassium cyanide will remain unreacted.
Now assume that $x{\text{ moles}}$ of silver ions are obtained due to the dissociation of the complex ion ${\left[ {{\text{Ag}}{{\left( {{\text{CN}}} \right)}_{\text{2}}}} \right]^ - }$
$\left( {0.03 - x} \right){\text{ moles}}$ of silver ions are obtained due to the dissociation of the complex ion ${\left[ {{\text{Ag}}{{\left( {{\text{CN}}} \right)}_{\text{2}}}} \right]^ - }$ remain undissociated. $\left( {0.04 + 2x} \right){\text{ moles}}$ of cyanide ions will be formed.
Write an expression for the equilibrium constant and substitute values in the expression:

$$K = \dfrac{{\left[ {{\text{A}}{{\text{g}}^ + }} \right] \times {{\left[ {{\text{C}}{{\text{N}}^ - }} \right]}^2}}}{{{{\left[ {{\text{Ag}}{{\left( {{\text{CN}}} \right)}_{\text{2}}}} \right]}^ - }}} \\\ \Rightarrow 4 \times {10^{ - 19}} = \dfrac{{x \times {{\left( {0.04 + 2x} \right)}^2}}}{{{\text{0}}{\text{.03}} - {\text{x}}}} \\\$$

Since the value of the equilibrium constant is very small, approximate ${\text{0}}{\text{.03}} - {\text{x}}$ to 0.03 and $0.04 + 2x$ to 0.04.

$$4 \times {10^{ - 19}} = \dfrac{{x \times {{\left( {0.04} \right)}^2}}}{{{\text{0}}{\text{.03}}}} \\\ \Rightarrow x = 4 \times {10^{ - 19}} \times \dfrac{{0.03}}{{{{\left( {0.04} \right)}^2}}} \\\ \Rightarrow x = 7.5 \times {10^{ - 18}} \\\$$

Hence, the silver ion concentration in the solution is $7.50{\text{ }} \times {\text{ }}{10^{ - 18}}M$.

Hence, the correct option is the option (B).

Note: The approximation is valid only if the value of equilibrium constant is less than five percent of the value of x. If it is not so, then you cannot make the simplification using the approximation. In such a case, you need to solve the polynomial equation.