##### J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 16.103

As we will see in Chapter 20, most ordinary soaps are water-soluble sodium or potassium salts of long-chain fatty acids. However, soaps of divalent cations, such as Ca2+, are only slightly water-soluble 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[CH3(CH2)14COO]2 as 0.003 g/100 mL at 25 oC.

(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 Ca2+, 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 Ksp of the calcium palmitate. Then, convert the calcium ion concentration into molarity units. Next, calculate the reaction quotient and compare to Ksp. 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 Ksp:
 Ca[CH3(CH2)14COO]2(s) Ca2+(aq) + 2 CH3(CH2)14COO–(aq) Ksp = [Ca2+]e[CH3(CH2)14COO–]e2 Initial 0 0 Change + x + 2x Equilibrium x 2x

The variable x must be found from the solubility:

The molar mass of Ca[CH3(CH2)14COO]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

Ksp = [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 Ca2+ = 25×10–3 g/L = (25×10–3 g/L)/(40.1 g/mol) = 6.2×10–4 M

Calculate the reaction quotient:

Q = [Ca2+]init[CH3(CH2)14COO]2init = [6.2×10–4][0.10]2 = 6.2×10–6 > Ksp

Since Q > Ksp 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:
 Ca[CH3(CH2)14COO]2(s) Ca2+(aq) + 2 CH3(CH2)14COO–(aq) Ksp = [Ca2+]e[CH3(CH2)14COO–]e2 Initial 6.2×10–4 0.10 Change – x – 2x Equilibrium 6.2×10–4 – x 0.10 – 2x

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 Ca2+ 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