Question

If the solubility of calcium sulphate in water is $1.03\times {10}^{-5}$, what will be its solubility in $0.005M$ solution of sulphuric acid?

• A. $1.2\times {10}^{-8}mol{L}^{-1}$
• B. $2.1\times {10}^{-8}mol{L}^{-1}$
• C. $3.8\times {10}^{-8}mol{L}^{-1}$
• D. $7.2\times {10}^{-8}mol{L}^{-1}$

Answer 1

The correct option is B

$2.1\times {10}^{-8}mol{L}^{-1}$

$CaS{O}_4\left(s\right)\rightleftharpoons C{a}^{2+}\left(aq\right)+S{O}_4^{2-}\left(aq\right)$

Let the solubility be $s$

$K_{sp}$ = $s.s$ = $s^2$

$K_{sp}$ = $1.06\times {10}^{-10}$

Also pay attention to the concentrations of the salt $NaCl$ in the previous problem and the concentration of sulphuric acid in this problem! It was $0.1MNaCl$ and here we have $0.005M{H}_{2}S{O}_4$.

So it means that we can't use the previous approximation. Instead, we know that the solubility is usually very small for these (sparingly soluble) salts.

But $K_{sp}$ is a constant that doesn't change as long as the temperature is constant. (in ionic equilibrium, unless otherwise specified, temperature = 298 K)

So $K_{sp}$ = $s.(s+0.005)$ = $\left[C{a}^{2+}\right]\left[\right[C{a}^{2+}]+0.005]$ = $x(x+0.005)$ = $x^2+x.0.005$

If $x$ is very small, we might neglect $x^2$

So $x$ = $\frac{K_{sp}}{0.005}$ = $2.1\times {10}^{-8}mol{L}^{-1}$