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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.

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Presentation on theme: "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."— Presentation transcript:

1 Lecture 14

2 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 - ]

3 Mass Balance Don’t use activities (this isn’t equilibrium) Conservation of mass Put in 0.5M H 2 CO 3 0.5 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- ])

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

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

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

7 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

8 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 2+ + 2OH - (3) K sp = [Mg 2+ ][OH - ] 2 = 7.1 x 10 -12 2H 2 O ⇌ H 3 O + + OH - (4) K w = [H 3 O + ][OH - ] = 1.0 x 10 -14

9 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 10 -12 (4) K w = [H 3 O + ][OH - ] = 1.0 x 10 -14

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

11 6. Solve (cont.) Set [Mg 2+ ] = X, so [OH - ] = 2X Use (3) (X)(2X) 2 = 7.1 x 10 -12 [Mg 2+ ] = X = 1.2 x 10 -4 M

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

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


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