Lecture 14

Charge balance Sum of positive charges = sum of negative charges In natural waters: [H + ]+2[Ca 2+ ]+2[Mg 2+ ]+[Na + ]+[K + ]=[HCO 3 - ]+2[CO 3 2- ]+2[SO 4 2- ]+[Cl - ]+[NO 3 - ]+[OH - ] Don’t use activities (this isn’t equilibrium) Don’t forget [H + ] and [OH - ]

Mass Balance Don’t use activities (this isn’t equilibrium) Conservation of mass Put in 0.5M H 2 CO M = [H 2 CO 3 ]+[HCO 3 - ]+[CO 3 2- ] Put in unknown concentration of Na 2 CO 3 [Na + ] = 2[CO 3 2- ] [Na + ] = 2([H 2 CO 3 ]+[HCO 3 - ]+[CO 3 2- ])

Solving chemical equilibrium problems Solubility of Mg(OH) 2 in water?

1. Write a set of balanced chemical equations Mg(OH) 2 (s) ⇌ Mg OH - 2H 2 O ⇌ H 3 O + + OH - (in all aqueous reactions) Solubility of Mg(OH) 2 based on [Mg 2+ ]

2. Charge Balance Always only 1 equation. (1) 2[Mg 2+ ] + [H 3 O + ] = [OH - ]

3. Mass Balance [OH - ] total = [OH - ] from Mg(OH)2 + [OH - ] fromH2O (2) [OH - ] = 2[Mg 2+ ] + [H 3 O + ] Same as charge balance this time, but not always

4. Write equilibrium constants for all the chemical equations. though Activity comes in here, though it is normally ignored unless working in a moderately-high ionic strength system Mg(OH) 2 (s) ⇌ Mg OH - (3) K sp = [Mg 2+ ][OH - ] 2 = 7.1 x H 2 O ⇌ H 3 O + + OH - (4) K w = [H 3 O + ][OH - ] = 1.0 x

5. Count equations and unknowns Need as many equations as unknowns 3 unknowns Mg 2+, OH -, H 3 O + 3 independent equations (1) and (2) 2[Mg 2+ ] + [H 3 O + ] = [OH - ] (3) K sp = [Mg 2+ ][OH - ] 2 = 7.1 x (4) K w = [H 3 O + ][OH - ] = 1.0 x

6. Solve Simplify (1) because [Mg 2+ ] >> [H 3 O + ] So... [OH - ] = 2[Mg 2+ ] Set [Mg 2+ ] = X, so [OH - ] = 2X

6. Solve (cont.) Set [Mg 2+ ] = X, so [OH - ] = 2X Use (3) (X)(2X) 2 = 7.1 x [Mg 2+ ] = X = 1.2 x M

To do at home: Calculate ionic strength and activities Was it OK to disregard activities?

Calculate the molar solubility of calcium oxalate in a solution that has been buffered so that its pH = 4.0.