1. Chem 1310: Introduction to physical chemistry Part 3: Equilibria Miscellaneous

2. Solubility equilibrium Suppose we have CaCO 3(s) ⇋ Ca 2+ (aq) + CO 3 2- (aq) K = [Ca 2+ ][CO 3 2- ] = 3.8*10 -9 How much CaCO 3 will dissolve in 1L of water? Ca 2+ CO 3 2- I00 C+x+x+x+x Exx

3. Solubility equilibrium K = [Ca 2+ ][CO 3 2- ] = x 2 = 3.8*10 -9 x = 6.2*10 -5 mol/L Ca 2+ CO 3 2- I00 C+x+x+x+x Exx

4. More than one equilibrium Suppose we have CaCO 3(s) ⇋ Ca 2+ (aq) + CO 3 2- (aq) K 1 = [Ca 2+ ][CO 3 2- ] = 3.8*10 -9 CO 3 2- (aq) + H 2 O ⇋ HCO 3 - (aq) + OH - (aq) How much CaCO 3 will now dissolve in 1L of water?

5. More than one equilibrium Wrong approach: why? Ca 2+ CO 3 2- I00 C+x+x+x+x Exx

6. More than one equilibrium Correct approach: K 1 = x(x-y) = 3.8*10 -9 K 2 = y 2 /(x-y) = 2.1*10 -4 Ca 2+ CO 3 2- HCO 3 - OH - I000  0) C1+x+x+x+x-- C2-y-y+y+y+y+y Exx-yx-yyy

7. More than one equilibrium K 1 = x(x-y) = 3.8*10 -9 K 2 = y 2 /(x-y) = 2.1*10 -4 After some manipulation, we get y 3 (y+K 2 ) = K 1 K 2 2 If we assume y « K 2, we obtain y 3 = K 1 K 2 = 8.0*10 -13  y = 9.3*10 -5 mol/L  x = 1.3*10 -4 mol/L (c.f. 6.2*10 -5 mol/L before) But note, the y « K 2, approximation is not very good!