What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K?

Question:

What is the minimum volume of water required to dissolve $1 \mathrm{~g}$ of calcium sulphate at $298 \mathrm{~K}$ ? $\left(\right.$ For calcium sulphate, $K_{\text {sp }}$ is $\left.9.1 \times 10^{-6}\right)$.

 

Solution:

$\mathrm{CaSO}_{4(s)} \longleftrightarrow \mathrm{Ca}^{2+}{ }_{(a q)}+\mathrm{SO}_{4}^{2-}(a q)$

$K_{s p}=\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{SO}_{4}^{2-}\right]$

Let the solubility of $\mathrm{CaSO}_{4}$ be $s$.

Then, $K_{s p}=s^{2}$

$9.1 \times 10^{-6}=s^{2}$

$s=3.02 \times 10^{-5} \mathrm{~mol} / \mathrm{L}$

Molecular mass of $\mathrm{CaSO}_{4}=136 \mathrm{~g} / \mathrm{mol}$

Solubility of $\mathrm{CaSO}_{4}$ in gram $/ \mathrm{L}=3.02 \times 10^{-3} \times 136$

$=0.41 \mathrm{~g} / \mathrm{L}$

This means that we need $1 \mathrm{~L}$ of water to dissolve $0.41 \mathrm{~g}$ of $\mathrm{CaSO}_{4}$

Therefore, to dissolve $1 \mathrm{~g}$ of $\mathrm{CaSO}_{4}$ we require $=\frac{1}{0.41} \mathrm{~L}=2.44 \mathrm{~L}$ of water.

 

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