Question

The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The molar solubility of oxygen in water is $······\times {10}^{-5}\mathrm{mol}{\mathrm{dm}}^{-3}$(Round off to the Nearest Integer). Given: Henry’s law constant $={\mathrm{K}}_{\mathrm{H}}=8.0\times {10}^4\mathrm{kPa}$ for ${\mathrm{O}}_2$. The density of water with dissolved oxygen = $1.0\mathrm{kg}{\mathrm{dm}}^{-3}$

Answer 1

Step 1: Given data

• Pressure$\left(\mathrm{P}\right)=20\mathrm{kPa}$
• Henry’s law constant$\left({\mathrm{K}}_{\mathrm{H}}\right)=8.0\times {10}^4\mathrm{kPa}$

Step 2: Formula used

• We know that ${\mathrm{P}}_{\left(\mathrm{g}\right)}=\left[{\mathrm{K}}_{\mathrm{H}}\right]\mathrm{X}$
• Where ${\mathrm{P}}_{\left(\mathrm{g}\right)}=$Pressure, $\left[{\mathrm{K}}_{\mathrm{H}}\right]=$Henry Constant and $\mathrm{X}=$Solubility

Step 3: Compute Solubility

• Substitute the known values in the formula,
• $\begin{array}{rcl}20\times {10}^3& =& 8·0\times {10}^3\times {10}^4\times \text{Solubility}\\ \mathrm{Solubility}& =& \frac{20\times {10}^3}{8.0\times {10}^3\times {10}^4}\\ & =& 25\times {10}^{-5}\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{dm}}^3$}\right.\end{array}$

Hence, the solubility is $25{\times 10}^{-5}\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{dm}}^3$}\right.$.