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As we will see in Chapter 20, most ordinary soaps are watersoluble sodium or potassium salts of longchain fatty acids. However, soaps of divalent cations, such as Ca^{2+}, are only slightly watersoluble and are often seen in the common soap scum formed in hard water. A handbook lists the solubility of the typical calcium soap, calcium palmitate, Ca[CH_{3}(CH_{2})_{14}COO]_{2} as 0.003 g/100 mL at 25 ^{o}C.
(a) If sufficient sodium soap is used to produce a concentration of palmitate ion equal to 0.10 M in a water sample having 25 ppm Ca^{2+}, would you expect any soap scum to form?
(b) If a soap scum does form under the conditions in part (a), how many grams of calcium palmitate would precipitate in a bowl containing 6.5 L of the water?
First: use the solubility to find the K_{sp} of the calcium palmitate. Then, convert the calcium ion concentration into molarity units. Next, calculate the reaction quotient and compare to K_{sp}. Finally, if precipitation occurs (a likelihood, given the last part of the question) use the equilibrium conditions to find the amount of calcium palmitate precipitated from 6.5 L of hard water.
Find K_{sp}:








Initial 



Change 



Equilibrium 


The variable x must be found from the solubility:
The molar mass of Ca[CH_{3}(CH_{2})_{14}COO]_{2} = 40.1 + 32(12.0) + 62(1.0) + 4(16.0) = 550.1 g/mol
x = (0.003 g/100 mL)(1000 mL/1 L)/(550.1 g/mol) = 5.5×10^{–5} M
K_{sp} = [x][2x]^{2} = [5.5×10^{–5}][2×5.5×10^{–5}]^{2} = 6.7×10^{–13}
To answer part (a), convert the calcium concentration to molarity:
25 ppm Ca^{2+} = 25×10^{–3} g/L = (25×10^{–3} g/L)/(40.1 g/mol) = 6.2×10^{–4} M
Calculate the reaction quotient:
Q = [Ca^{2+}]_{init}[CH_{3}(CH_{2})_{14}COO^{–}]^{2}_{init} = [6.2×10^{–4}][0.10]^{2} = 6.2×10^{–6} > K_{sp}
Since Q > K_{sp} there too many ions in solution to satisfy equilibrium and precipitation must occur to reach equilibrium.
For part (b), to find the amount of precipitate:








Initial 



Change 



Equilibrium 


Since x < 6.2×10^{–4} (the reaction must stop when the calcium ion is gone), the approximation 0.10 – 2x ~ 0.10 is valid. However, x may not be small compared to 6.2×10^{–4} so
6.7×10^{–13} = [6.2×10^{–4} – x][0.10]^{2}
x = 6.2×10^{–4} M = the molarity of calcium ions that precipitated from solution to give calcium palmitate.
To find the number of moles of calcium ion that precipited:
moles Ca^{2+} precipitated = (6.2×10^{–4} mol/L)(6.5 L) = 4.0×10^{–3} mol
The stoichiometric ratio between calcium ions and calcium palmitate is 1:1, so the number of moles of calcium palmitate precipitated is 4.0×10^{–3} mol.
Finally, convert this to grams: (4.0×10^{–3} mol)(550.1 g/mol) = 2.2 g