Animated Solution for Chemistry - Ionic Equilibrium: The solubility of $\text{AgCN}$ in a buffer solution of $\text{pH}=3$ is $x$. The value of $x$ is...... . [Assume : No cyano complex is formed; $K_{sp}(\text{AgCN})=2.2\times 10^{-16}$ and $K_a(\text{HCN})=6.2\times 10^{-10}$ ]

How pH Controls Solubility

Imagine you are standing in a laboratory, holding a small block of solid silver cyanide ($\text{AgCN}$). You drop it into a beaker containing pure water. It barely dissolves. The solubility product, $K_{sp}$, is a minuscule $2.2\times 10^{-16}$. It seems like a lost cause.

But what if we change the rules of the game? What if, instead of pure water, we drop it into a highly acidic buffer solution maintained strictly at a pH of 3? Suddenly, the chemistry becomes a thrilling tug-of-war between two simultaneous equilibria.

When $\text{AgCN}$ enters the solution, it attempts to establish its standard solubility equilibrium:

The equilibrium constant for this process is the solubility product, $K_{sp}$. If this were the only reaction happening, the solubility $S$ would simply be $\sqrt{K_{sp}}$. But in an acidic buffer, the ${\text{CN}}^{-}$ ions are not safe.

The cyanide ion (${\text{CN}}^{-}$) is the conjugate base of a weak acid, hydrocyanic acid ($\text{HCN}$). Because the buffer maintains a high concentration of protons ($[{\text{H}}^{+}]=10^{-3}\text{ M}$), these protons aggressively attack the newly formed ${\text{CN}}^{-}$ ions:

The equilibrium constant for this protonation is the reciprocal of the acid dissociation constant, $1/K_a$. Since $K_a$ is very small ($6.2\times 10^{-10}$), $1/K_a$ is massive. This means almost every single ${\text{CN}}^{-}$ ion that manages to break free from the solid $\text{AgCN}$ is immediately converted into $\text{HCN}$.

According to Le Chatelier's Principle, as ${\text{CN}}^{-}$ is continuously removed from the primary equilibrium, the system is forced to dissolve more solid $\text{AgCN}$ to compensate. The acid is literally pulling the solid apart!

To find the true solubility, we must combine these two reactions into a single net ionic equation:

When we add chemical equations, we multiply their equilibrium constants. Therefore, the net equilibrium constant $K_{eq}$ is:

Let the new solubility of the salt be $S$. For every mole of $\text{AgCN}$ that dissolves, one mole of ${\text{Ag}}^{+}$ is produced, so $[{\text{Ag}}^{+}]=S$. Because the conversion of ${\text{CN}}^{-}$ to $\text{HCN}$ is nearly 100% complete in this acidic environment, we can safely approximate that $[\text{HCN}]\approx S$.

Now, we substitute our variables into the master equation. The buffer ensures that $[{\text{H}}^{+}]$ remains constant at $10^{-3}\text{ M}$:

Rearranging to solve for $S^2$:

To make taking the square root easier, we adjust the decimal point to create an even power of 10:

Finally, taking the square root yields the solubility:

Rounding to the nearest given option, we get $1.9\times 10^{-5}\text{ M}$.

By lowering the pH, we increased the solubility of $\text{AgCN}$ by several orders of magnitude compared to pure water. This beautiful interplay of simultaneous equilibria is a cornerstone of analytical chemistry and a favorite concept in competitive exams!