The acid ionization of hydrated aluminium ion is<br><br><br><br>How many milligrams of should be dis — Ionic Equilibrium Chemistry Question
Question
The acid ionization of hydrated aluminium ion is<br><br>$\text{Al(H}_2\text{O)}_6^{3+}\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{Al(H}_2\text{O)}_5\text{OH}^{2+}\text{(aq)} + \text{H}_3\text{O}^+\text{(aq)}; K_a = 1.0 \times 10^{-5}$<br><br>How many milligrams of $\text{AlCl}_3$ should be dissolved in sufficient water to get 400 ml of solution of pH, 3.0?
💡 Solution & Explanation
Given $\text{pH} = 3.0$, $[\text{H}_3\text{O}^+] = 10^{-3}\text{ M}$. From the stoichiometry, $[\text{Al(H}_2\text{O)}_5\text{OH}^{2+}] = 10^{-3}\text{ M}$. Using $K_a = \frac{(10^{-3})^2}{[\text{Al}^{3+}]_{\text{eq}}} = 1.0 \times 10^{-5} \implies [\text{Al}^{3+}]_{\text{eq}} = 0.1\text{ M}$. Total initial concentration $C = 0.1 + 10^{-3} = 0.101\text{ M}$. In most textbook approximations for weak acids, if $C \approx 0.1\text{ M}$, moles in $400\text{ ml}$ is $0.1 \times 0.4 = 0.04\text{ moles}$. Molar mass of $\text{AlCl}_3 = 27 + 3(35.5) = 133.5\text{ g/mol}$. Mass required = $0.04 \times 133.5 = 5.34\text{ g} = 5340\text{ mg}$. Formatted to four digits, this is 5340.