What mass of NaOH should be dissolved in sufficient water to get of an aqueous solution of pH, 7.3, — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

$\text{pH} = 7.3 \implies [\text{H}^+] = 10^{-7.3} = 10^{0.7} \times 10^{-8} \approx 5.0 \times 10^{-8}\text{ M}$. The corresponding $[\text{OH}^-] = 10^{-14} / (5 \times 10^{-8}) = 2.0 \times 10^{-7}\text{ M}$. The total $[\text{OH}^-]$ comes from NaOH and the autoionization of water. By charge balance: $[\text{Na}^+] + [\text{H}^+] = [\text{OH}^-]$, so $[\text{Na}^+] = 2.0 \times 10^{-7} - 5.0 \times 10^{-8} = 1.5 \times 10^{-7}\text{ M}$. This is the concentration of added NaOH. Volume = $20\text{ m}^3 = 20,000\text{ L}$. Moles of NaOH = $1.5 \times 10^{-7}\text{ mol/L} \times 20,000\text{ L} = 3.0 \times 10^{-3}\text{ moles}$. Mass of NaOH = $3.0 \times 10^{-3}\text{ mol} \times 40\text{ g/mol} = 0.12\text{ g}$.