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 mol L⁻¹ to 10⁻³ mol L⁻¹. The pK_a of HX is

The correct option is 4.

Step 1: Understand the solubility equilibrium of a sparingly soluble salt of a weak acid
Consider a sparingly soluble salt MX dissolving in an aqueous solution. It dissociates into M⁺ and X^- ions:

$\text{MX}\left(\text{s}\right)\rightleftharpoons {\text{M}}^{+}\left(\text{aq}\right)+{\text{X}}^{-}\left(\text{aq}\right)$

The solubility product constant ${\text{K}}_{\text{sp}}$ is given by:

${\text{K}}_{\text{sp}}=\left[{\text{M}}^{+}\right]\left[{\text{X}}^{-}\right]$

Since HX is a weak acid, the anion X^- reacts with H⁺ ions present in the solution to form HX:

${\text{X}}^{-}\left(\text{aq}\right)+{\text{H}}^{+}\left(\text{aq}\right)\rightleftharpoons \text{HX}\left(\text{aq}\right)$

The acid dissociation constant ${\text{K}}_{\text{a}}$ of the acid HX is given by:

${\text{K}}_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{X}}^{-}\right]}{\left[\text{HX}\right]}$

From this, the concentration of unprotonated anion [X^-] can be expressed in terms of [HX] as:

$\left[\text{HX}\right]=\frac{\left[{\text{H}}^{+}\right]\left[{\text{X}}^{-}\right]}{{\text{K}}_{\text{a}}}$

Step 2: Express molar solubility (S) in terms of concentrations
Let S be the molar solubility of the salt MX at a given pH.
The total concentration of dissolved M⁺ ions is:

$\left[{\text{M}}^{+}\right]=\text{S}$

The total dissolved X species exists as both free X^- and protonated HX:

$\text{S}=\left[{\text{X}}^{-}\right]+\left[\text{HX}\right]$

Substitute [HX] into the equation:

$\text{S}=\left[{\text{X}}^{-}\right]+\frac{\left[{\text{H}}^{+}\right]\left[{\text{X}}^{-}\right]}{{\text{K}}_{\text{a}}}=\left[{\text{X}}^{-}\right](1+\frac{\left[{\text{H}}^{+}\right]}{{\text{K}}_{\text{a}}})$

Therefore, the concentration of free anion [X^-] is:

$\left[{\text{X}}^{-}\right]=\frac{\text{S}}{1+\frac{\left[{\text{H}}^{+}\right]}{{\text{K}}_{\text{a}}}}$

Now, substitute [M⁺] = S and [X^-] into the expression for ${\text{K}}_{\text{sp}}$:

${\text{K}}_{\text{sp}}=\text{S}·\frac{\text{S}}{1+\frac{\left[{\text{H}}^{+}\right]}{{\text{K}}_{\text{a}}}}=\frac{{\text{S}}^2}{1+\frac{\left[{\text{H}}^{+}\right]}{{\text{K}}_{\text{a}}}}$

Rearranging for ${\text{S}}^2$:

${\text{S}}^2={\text{K}}_{\text{sp}}(1+\frac{\left[{\text{H}}^{+}\right]}{{\text{K}}_{\text{a}}})$

Step 3: Set up equations using the given data
At pH = 7:
$[{\text{H}}^{+}{]}_1={10}^{-7}\text{ M}$ and solubility ${\text{S}}_1={10}^{-4}\text{ mol L}{}^{-1}$

$({10}^{-4}{)}^2={\text{K}}_{\text{sp}}(1+\frac{{10}^{-7}}{{\text{K}}_{\text{a}}})$ --- (Equation 1)

At pH = 2:
$[{\text{H}}^{+}{]}_2={10}^{-2}\text{ M}$ and solubility ${\text{S}}_2={10}^{-3}\text{ mol L}{}^{-1}$

$({10}^{-3}{)}^2={\text{K}}_{\text{sp}}(1+\frac{{10}^{-2}}{{\text{K}}_{\text{a}}})$ --- (Equation 2)

Step 4: Solve for ${\text{K}}_{\text{a}}$ and ${\text{pK}}_{\text{a}}$
Divide Equation 2 by Equation 1:

$\frac{({10}^{-3}{)}^2}{({10}^{-4}{)}^2}=\frac{1+\frac{{10}^{-2}}{{\text{K}}_{\text{a}}}}{1+\frac{{10}^{-7}}{{\text{K}}_{\text{a}}}}$

$\frac{{10}^{-6}}{{10}^{-8}}=100=\frac{1+\frac{{10}^{-2}}{{\text{K}}_{\text{a}}}}{1+\frac{{10}^{-7}}{{\text{K}}_{\text{a}}}}$

Since ${\text{K}}_{\text{a}}\gg {10}^{-7}$ for standard weak acid solutions in this range, $\frac{{10}^{-7}}{{\text{K}}_{\text{a}}}\ll 1$, so $1+\frac{{10}^{-7}}{{\text{K}}_{\text{a}}}\approx 1$.

Thus, the equation simplifies to:

$100=1+\frac{{10}^{-2}}{{\text{K}}_{\text{a}}}$

$\frac{{10}^{-2}}{{\text{K}}_{\text{a}}}=99\approx {10}^2$

${\text{K}}_{\text{a}}=\frac{{10}^{-2}}{{10}^2}={10}^{-4}$

Finally, calculate the ${\text{pK}}_{\text{a}}$:

${\text{pK}}_{\text{a}}=-{\text{log}}_{10}\left({\text{K}}_{\text{a}}\right)=-{\text{log}}_{10}\left({10}^{-4}\right)=4$

Hence, the pK_a of HX is equal to 4.