The solubility of a salt of a weak acid $MX$ (e.g.,phosphoric acid salts) increases at lower $pH$. Explain with equations.

(A) The solubility of a salt of a weak acid like $Na_{3}PO_{4}$ increases at lower $pH$ because at lower $pH$,the concentration of the anion $X^{-}$ decreases due to protonation. As $[X^{-}]$ decreases,the solubility of $MX$ increases to maintain the solubility product constant.
$MX_{(s)} \rightleftharpoons M_{(aq)}^{+} + X_{(aq)}^{-} \quad \dots (Eq.-i)$
$K_{sp} = [M^{+}][X^{-}] \quad \dots (Eq.-ii)$
Since $MX$ is the salt of a weak acid $(HX)$,the ionization of the weak acid is as follows:
$HX_{(aq)} \rightleftharpoons H_{(aq)}^{+} + X_{(aq)}^{-} \quad \dots (Eq.-iii)$
Note: In this salt and weak acid,the common ion is $X^{-}$.
The ionization constant for the weak acid is:
$K_{a} = \frac{[H^{+}][X^{-}]}{[HX]} \quad \dots (Eq.-iv)$
From $(Eq.-iv)$,we get $\frac{[X^{-}]}{[HX]} = \frac{K_{a}}{[H^{+}]}$.
Let $f$ be the fraction of the anion $X^{-}$ present in the solution:
$f = \frac{[X^{-}]}{[HX] + [X^{-}]} = \frac{K_{a}}{[H^{+}] + K_{a}} \quad \dots (Eq.-v)$
As $[H^{+}]$ increases,$pH$ decreases,and the value of '$f$' decreases,which shifts the equilibrium of $(Eq.-i)$ to the right,thereby increasing the solubility.