Fe(OH)2 (S)  Fe2+ (aq) + 2 OH- (aq)

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Presentation transcript:

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 * 10-16 = [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 * 10-16 = [Fe++](1 * 10-4) X as [Fe++] = 8 * 10-8 M = molar solubility

CH 17 # 66 Part C: pH+ = 12.0, pOH- = 2.0 : [OH-] = 1 * 10-2 Ksp = [Fe++][OH-] Ksp = 7.9 * 10-16 = [Fe++](1 * 10-2) X as [Fe++] = 8 * 10-12 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 * 10-10 = [0.010][SO42-] X as [SO42-] = 1.1 * 10-8 M = molar concentration of sulfate needed at saturation.

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 .