Most ordinary soaps are sodium salt of long chain fatty acids and are soluble in water. Soaps of div — Ionic Equilibrium Chemistry Question
Question
Most ordinary soaps are sodium salt of long chain fatty acids and are soluble in water. Soaps of divalent cations such as $\text{Ca}^{2+}$ are only slightly soluble and are often seen in the common soap is calcium palmitate, $\text{Ca[CH}_3\text{(CH}_2\text{)}_{14}\text{COO]}_2$. A handbook of chemistry lists the solubility of this soap as 0.0055 g per 100 ml at $25^\circ\text{C}$. If sufficient sodium soap is used to produce a final concentration of palmitate ion equal to 0.10 M in a water sample having 40 ppm $\text{Ca}^{2+}$ initially, how many milligrams of calcium palmitate would precipitate in a bowl containing 10 l of this water sample?
💡 Solution & Explanation
Molar mass of calcium palmitate: Palmitate ($\text{C}_{15}\text{H}_{31}\text{COO}^-$) is 255. Calcium palmitate ($\text{Ca(Pal)}_2$) is $40 + 2(255) = 550\text{ g/mol}$. Solubility $S = 0.0055\text{ g} / 100\text{ ml} = 0.055\text{ g/L}$. Molar solubility $S = 0.055 / 550 = 10^{-4}\text{ M}$. $K_{sp} = [\text{Ca}^{2+}][\text{Pal}^-]^2 = (10^{-4})(2 \times 10^{-4})^2 = 4 \times 10^{-12}$. Initial $[\text{Ca}^{2+}] = 40\text{ ppm} = 40\text{ mg/L} = 0.04\text{ g/L} \implies 0.04 / 40 = 10^{-3}\text{ M}$. If final $[\text{Pal}^-] = 0.10\text{ M}$, the equilibrium $[\text{Ca}^{2+}] = \frac{K_{sp}}{(0.10)^2} = 4 \times 10^{-10}\text{ M}$, meaning practically all $\text{Ca}^{2+}$ precipitates. Moles of $\text{Ca}^{2+}$ precipitated in 1 L is $10^{-3}\text{ mol}$. In a 10 L bowl, $10^{-2}\text{ moles}$ precipitate. Mass = $10^{-2} \times 550 = 5.5\text{ g} = 5500\text{ mg}$. Formatted to four digits, this is 5500.