1. Chapter 16 Solubility Equilibria

2. Saturated solutions of “insoluble” salts are another type of chemical equilibria. Ionic compounds that are termed “insoluble” actually have a low, but definite solubility. For a saturated solution of AgCl, the equation would be: AgCl(s)  Ag + (aq) + Cl  (aq) The solubility product expression would be: K sp = [Ag + ][Cl  ] The AgCl(s) is left out since solids are left out of equilibrium expressions (constant concentrations). For Ag 2 CO 3, Ag 2 CO 3  2Ag + + CO 3 2  K sp = [Ag + ] 2 [CO 3 2  ]

3. K sp = Solubility Product Constant The K sp of a substance is a measure of its solubility. K sp values cannot be directly compared if the number of ions in the formula differs. Smaller K sp values indicate lower solubility.

4. The K sp of AgCl is 1.6×10  10. This means that if the product of [Ag + ][Cl - ] 1.6×10  10, the solution is saturated and a solid (precipitate) would form. The product of the ions (raised to the power of their coefficients) is called the ion product quotient or Q. If Q > K sp, ppt forms. If Q < K sp, no ppt forms.

5. 4 Main Types of Problems 1. Given molar solubility, find K sp. 2. Given K sp, find molar solubility. 3. Given K sp, find molar solubility in the presence of a common ion. 4. Will a precipitate form when these two solutions are mixed? Let’s look at an example of each.

6. Ex. The molar solubility of silver sulfate is 1.5×10  2 mol/L. Calculate the solubility product of the salt. Reaction Ag 2 SO 4 (s)  2Ag + + SO 4 2  Initial ---- 0 0 Change -x +2x +x Equil. ---- 3.0×10  2 1.5×10  2 x = 1.5×10  2 Since 1.5×10  2 mol/L of Ag 2 SO 4 dissolve, 1.5×10  2 mol/L of SO 4 2- form and 2(1.5×10  2 mol/L) of Ag + form. K sp = [Ag + ] 2 [SO 4 2  ] = (3.0×10  2 ) 2 (1.5×10  2 )= 1.4 × 10  5 Remember that molar solubility is “x”!

7. Ex. Calculate the molar solubility of calcium phosphate. The K sp of calcium phosphate is 1.2×10  26. Reaction Ca 3 (PO 4 ) 2  3Ca 2+ + 2PO 4 3  Initial --- 0 0 Change  x +3x +2x Equilibrium --- 3x 2x K sp = [Ca 2+ ] 3 [PO 4 3  ] 2 1.2×10  26 = (3x) 3 (2x) 2 = 108x 5 x 5 = 1.1×10  28 x = 2.6×10  6 M

8. Ex. What is the molar solubility of lead(II) iodide in a 0.050 M solution of sodium iodide? (Common ion effect problem) Reaction PbI 2  Pb 2+ + 2I  Initial ---- 0 0.050M Change  x +x +2x Equil. ---- x 0.050 + 2x K sp = [Pb 2+ ][I  ] 2 K sp = 1.4×10  8 = (x)(0.050+2x) 2  x(0.050) 2 x = 5.6 × 10  6 M The molar solubility of PbI 2 in pure water is 1.5×10  3 M. This shows the decreased solubility of a salt in the presence of a common ion. Don’t forget to put in the initial concentration of the common ion!

9. Ex. Exactly 200 mL of 0.040 M BaCl 2 are added to exactly 600 mL of 0.080 M K 2 SO 4. Will a precipitate form? BaCl 2 + K 2 SO 4  BaSO 4 + 2KCl Barium sulfate is the likely precipitate. 0.200 L×0.040 M BaCl 2 = 8.0×10  3 mol Ba 2+ 8.0×10  3 mol/0.800L total volume = 1.0×10  2 M Ba 2+ 0.600 L×0.080 M K 2 SO 4 = 4.8×10  2 mol SO 4 2  4.8×10  2 mol/0.800L = 6.0×10  2 M SO 4 2  K sp = [Ba 2+ ][SO 4 2  ] = 1.1×10  10 (look this up in table) Q = (1.0×10  2 )(6.0×10  2 ) = 6.0×10  4 Q > K sp 6.0×10  4 > 1.1×10  10 A precipitate of BaSO 4 forms.

10. Shortcut Alert In a problem involving dilution (like the previous problem), the volumes are often equal. When the volumes are equal and then combined, concentrations are halved. For example, if 400 mL of each solution in the previous problem were combined, the resulting concentrations would be: [Ba 2+ ]= 0.0400/2 = 0.0200 M and [SO 4 2  ] = 0.0800/2 = 0.0400 M