A sparingly soluble salt MX has K_sp = 2.0x 10^-10. The anion X- is the conjugate base of the weak acid HX (K_a = 5.0x 10^-5). Find the solubility of MX in a buffer of pH 3.00 (assume constant pH; X present as X- and HX). Give s in units of 10^-5 M to two decimals.

JEE Main levelNumerical answerDifficulty 2 of 3ia154

A sparingly soluble salt MX has $K_{s p} = 2.0 \times 1 0^{- 10}$ . The anion $X X^{-}$ is the conjugate base of the weak acid HX ( $K_{a} = 5.0 \times 1 0^{- 5}$ ). Find the solubility of MX in a buffer of pH $3.00$ (assume constant pH; X present as $X X^{-}$ and HX). Give $s$ in units of $1 0^{- 5} M$ to two decimals.

Numerical answer type, answer in x 10^-5 M, tolerance 0.02.

s = [M X^{+}] = [X X^{-}] + [HX] = [X X^{-}] (1 + [H X^{+}] / K_{a})

K_{s p} = [M X^{+}] [X X^{-}] = s^{2} / (1 + [H X^{+}] / K_{a})

, so

s = K_{s p} (1 + 1 0^{- 3} /5.0 \times 1 0^{- 5}) = 2.0 \times 1 0^{- 10} \times 21 = 6.48 \times 1 0^{- 5} M

, i.e.

6.48 \times 1 0^{- 5} M