CHEMISTRY 31

Summer, 2016 - Dixon

Homework Additional Problem 2.2 Solutions

SrCO₃ is a sparingly soluble salt. Using the K_sp from Appendix F,

a) Determine the concentrations Sr²⁺ and CO₃^2- for a saturated SrCO₃ solution in pure water ignoring secondary reactions of Sr²⁺ and CO₃^2- and ignoring the activity from dissolved Sr²⁺ and CO₃^2-.

This can be solved the traditional way – by setting up an ICE table

SrCO₃ (s) ↔ Sr²⁺ + CO₃^2-

Initial 0 0

Change +x +x

Equilibrium x x

Now we can use the K for the reaction K_sp = 9.3 x 10^-10 = [Sr^2+][CO₃^2-] = x² or x = [Sr²⁺] = [CO₃^2-] = (9.3 x 10^-10) ^0.5 = 3.05 x 10^-5 M

b) Using the concentrations of Sr²⁺ and CO₃^2- found in a), determine the ionic strength of the solution, the activity coefficients of Sr²⁺ and CO₃^2-, and account for these in a recalculated concentration of Sr²⁺ and CO₃^2- (only one iteration of this calculated is expected). Continue to ignore reactions of Sr²⁺ and CO₃^2-.

m = 0.5[(3.05 x 10^-5)*4*2] = 1.22 x 10^-4 M

Now we can look up a(Sr²⁺) = 500 and a(CO₃^2-) = 450 and calculate g(Sr²⁺) and g(CO₃^2-)

Logg(Sr²⁺) = [(-0.51)(+2)²(1.22 x 10^-4)^0.5]/[1+500(1.22 x 10^-4)^0.5/305] = -0.023/1.018 = -0.022 so g(Sr²⁺) = 0.950

Using the same equation for CO₃^2-, logg(CO₃^2-) = -0.023/1.016 = -0.022 and g(CO₃^2-) = 0.950

Now, the ICE part will be the same, but we need to use the activity coefficient corrected equilibrium equation:

K_sp = 9.3 x 10^-10 = g(Sr^2+)[Sr²⁺]g(CO₃^2-)[CO₃^2-] = (0.950)²x² or x = (9.3 x 10^-10/0.950²)^0.5 = 3.2 x 10^-5 M

c) Now, using equilibrium constant values from Appendices G, I and J, write out other significant reactions (e.g.of Sr²⁺ or CO₃^2- with water or with each other) needed to solve this problem using the systematic method. For reactions from Appendix I and J with b or K values less than 10 (logb or logK values less than 1), you can ignore those reactions. Additionally, write out a charge balance equation.

Secondary reactions:

CO₃^2- + H₂O(l) ↔ HCO₃^- + OH^-

HCO₃^- + H₂O(l) ↔ H₂CO₃(aq) + OH^-

Sr²⁺ + CO₃^2- ↔ SrCO₃(aq) K = 10^2.81 > 10

H₂O(l) ↔ H⁺ + OH^-

Sr²⁺ + OH^- ↔ SrOH⁺ (should omit this because K < 10)

Charge balance equation: 2[Sr²⁺] + [H⁺] = 2[CO₃^2-] + [HCO₃^-] + [OH^-]