Animated Solution for Chemistry - Ionic Equilibrium: On decreasing the pH from 7 to 2, the solubility of a sparingly soluble salt (MX) of a weak acid (HX) increased from $10^{-4}{\text{ mol L}}^{-1}$ to $10^{-3}{\text{ mol L}}^{-1}$. The $\text{p}{K}_a$ of HX is:

Imagine a beaker with a sparingly soluble salt, $\text{MX}$, sitting quietly at the bottom. In pure water, it barely dissolves. The equilibrium is simple: $\text{MX(s)}\rightleftharpoons {\text{M}}^{+}(\text{aq})+{\text{X}}^{-}(\text{aq})$.

But what happens when we drastically drop the pH from 7 to 2? We are flooding the solution with ${\text{H}}^{+}$ ions. These protons are highly reactive and immediately seek out the ${\text{X}}^{-}$ ions to form the weak acid $\text{HX}$.

This is a classic, beautiful application of Le Chatelier's Principle. By continuously removing ${\text{X}}^{-}$ from the solution to form $\text{HX}$, the system feels a deficit. To compensate, it forces more solid $\text{MX}$ to dissolve. This is exactly why the solubility skyrockets from $10^{-4}\text{ M}$ to $10^{-3}\text{ M}$.

Let's establish our baseline. At pH 7, we are dealing with pure water. The solubility is given as $s_1=10^{-4}{\text{ mol L}}^{-1}$.

Since one mole of $\text{MX}$ yields one mole of ${\text{M}}^{+}$ and one mole of ${\text{X}}^{-}$, their concentrations are both equal to $s_1$.

The solubility product is simply $K_{sp}=[{\text{M}}^{+}][{\text{X}}^{-}]=s_1^2$. Taking the square root, we find that $\sqrt{K_{sp}}=s_1=10^{-4}$. We will keep this elegant relationship in our back pocket for later.

Simultaneous Equilibria

Now, let's analyze the chaotic state at pH 2. The concentration of ${\text{H}}^{+}$ is $10^{-2}\text{ M}$, and the new solubility $s$ is $10^{-3}\text{ M}$. We now have two competing equilibria happening simultaneously:

1. The dissolution of the salt: $\text{MX(s)}\rightleftharpoons {\text{M}}^{+}+{\text{X}}^{-}$ (Governed by $K_{sp}$) 2. The formation of the weak acid: ${\text{X}}^{-}+{\text{H}}^{+}\rightleftharpoons \text{HX}$ (Governed by $\frac{1}{K_a}$)

The total amount of ${\text{M}}^{+}$ in solution is exactly the new solubility, $s$. The total amount of the $\text{X}$ species is also $s$, but it is now split between the free ${\text{X}}^{-}$ ions and the protonated $\text{HX}$ molecules. If we let $[\text{HX}]=x$, then the remaining free $[{\text{X}}^{-}]=s-x$.

Here is where physical intuition saves us from a nightmare of quadratic equations. Because $\text{HX}$ is a weak acid and we have a massive excess of ${\text{H}}^{+}$ ($10^{-2}\text{ M}$ is huge compared to the solubility), the equilibrium ${\text{X}}^{-}+{\text{H}}^{+}\rightleftharpoons \text{HX}$ is driven violently to the right.

Almost all the dissolved ${\text{X}}^{-}$ gets converted into $\text{HX}$. Therefore, we can safely approximate that $x\approx s$. This means the concentration of $\text{HX}$ is roughly equal to the total solubility $s$.

Let's set up our two master equations with this approximation in mind:

From the solubility product:

From the acid dissociation constant $K_a=\frac{[{\text{H}}^{+}][{\text{X}}^{-}]}{[\text{HX}]}$:

Now, prepare for a brilliant mathematical trick. If we multiply Equation (1) by Equation (2), the annoying $(s-x)$ term perfectly cancels out!

Taking the square root of both sides gives us a pristine formula for the new solubility: $s=\frac{\sqrt{K_{sp}}}{10\sqrt{K_a}}$.

We are in the endgame. We know the ratio of the new solubility to the old solubility is $\frac{s}{s_1}=\frac{10^{-3}}{10^{-4}}=10$.

Substituting our algebraic expressions:

The $\sqrt{K_{sp}}$ terms beautifully cancel out, leaving:

Squaring both sides yields $K_a=10^{-4}$. Taking the negative logarithm, we arrive at our final, elegant answer: $\text{p}{K}_a=4$.