1. Fe(OH)2 (S)  Fe2+ (aq) + 2 OH- (aq)
CH 17 # 54
Fe(OH)2 (S)  Fe2+ (aq) + 2 OH- (aq)
Part A: pH+ = 7.0, pOH- = 7.0 : [OH-] = 1 * 10-7
Ksp = [Fe++][OH-]
Ksp = 7.9 * = [Fe++](1 * 10-7)
X as [Fe++] = 8 * 10-2 M = molar solubility
Part B: pH+ = 10.0, pOH- = 4.0 : [OH-] = 1 * 10-4
Ksp = [Fe++][OH-]
Ksp = 7.9 * = [Fe++](1 * 10-4)
X as [Fe++] = 8 * 10-8 M = molar solubility

2. CH 17 # 66 Part C: pH+ = 12.0, pOH- = 2.0 : [OH-] = 1 * 10-2
Ksp = [Fe++][OH-]
Ksp = 7.9 * = [Fe++](1 * 10-2)
X as [Fe++] = 8 * M = molar solubility
CH 17 # 66
Part A: Precipitation will begin when Ksp = ion prod **** Ksp = [Ba++] [SO42-] = 1.1*10-10
Ksp = 1.1 * = [0.010][SO42-]
X as [SO42-] = 1.1 * 10-8 M = molar concentration of sulfate needed at saturation.

3. CH 17 # 66 continued
Part A: Precipitation will begin when Ksp = ion prod **** Ksp = [Sr++] [SO42-] = 3.2*10-7
Ksp = 3.2 * 10-7 = [0.010][SO42-]
X as [SO42-] = 3.2 * 10-5 M = molar concentration of sulfate needed at saturation.
Ba++ precipitates first as it needs a lower molarity of sulfate to reach saturation.
Sr++ precipitates when the molarity of sulfate reaches
3.2 * 10-5 M .