Chapter 16, Problem 80AE

Interpretation Introduction

Interpretation:

Mass of ${\text{CaSO}}_{\text{4}}$ dissolved in $\text{600 mL}$ of water has to be calculated.

Concept Introduction:

The equilibrium constant used for the partially soluble salt in water is termed as solubility product constant $K_{\text{sp}}$. Consider solubility of $\text{AB}$ in water and equilibrium equation of $\text{AB}$ is as follows:

  $\text{AB}\left(s\right)\rightleftharpoons {\text{A}}^{+}\left(aq\right)+{\text{B}}^{-}\left(aq\right)$

The expression for $K_{\text{sp}}$ is as follows:

  $K_{\text{sp}}=\frac{\left[{\text{A}}^{+}\right]\left[{\text{B}}^{-}\right]}{\left[\text{AB}\left(s\right)\right]}$

Generally the concentration of solid is taken as constant. Therefore the expression for $K_{\text{sp}}$ of $\text{AB}$ is as follows:

  $K_{\text{sp}}=\left[{\text{A}}^{+}\right]\left[{\text{B}}^{-}\right]$

Concentration of solution is expressed in terms of molarity. It is the ratio of moles of solute to the volume of solution in litres. The expression used to determine the concentration of solution is as follows:

  $\text{Concentration of solution}=\frac{\text{moles of solute}}{\text{volume of solution}\left(\text{L}\right)}$

Mole is the unit that measures the quantity such as atoms, volume and molecules. It is the ratio of given mass to molecular mass of substance. It is expressed as follows:

$\text{Number of moles}=\frac{\text{given mass}}{\text{molecular mass}}$