Help ! I’m dissolving

Using complex algebra to calculate concentrations when multiple equilibria are in play

 We can look for two more pieces of info: Charge Balance: Electroneutrality of the solution; the sum of the positive charge in solution equals the sum of negative charges in the solution.

Mass Balance: Conservation of matter; the quantity of all species in a solution containing a particular atom (or group of atoms) must equal to the amount of that atom (or group) delivered to the solution.

 We can look for two more pieces of info: Charge Balance: Electroneutrality of the solution; the sum of the positive charge in solution equals the sum of negative charges in the solution. Mass Balance: Conservation of matter; the quantity of all species in a solution containing a particular atom (or group of atoms) must equal to the amount of that atom (or group) delivered to the solution. 11 [H + ] + 2[Ca 2+ ] = [OH - ] + [F - ](4) 2[Ca 2+ ] = [F - ] + [HF](5)

How does the solubility of CaF 2 depend on pH? 12 CaF 2 (s)  Ca F - K sp = [Ca 2+ ][F - ] 2 = 3.9  (1) F - +H 2 O  HF + OH - (2) H 2 O  H + + OH - K w = [H + ][OH - ] = 1.0  (3) Five unknowns: [Ca 2+ ], [F - ], [HF], [H + ], and [OH - ] Three equations mean we need more equations…

 We can look for two more pieces of info: Charge Balance: Electroneutrality of the solution; the sum of the positive charge in solution equals the sum of negative charges in the solution. Mass Balance: Conservation of matter; the quantity of all species in a solution containing a particular atom (or group of atoms) must equal to the amount of that atom (or group) delivered to the solution. [H + ] + 2[Ca 2+ ] = [OH - ] + [F - ](4) 2[Ca 2+ ] = [F - ] + [HF](5)

K sp = [Ca 2+ ][F - ] 2 = 3.9  (1) (2) K w = [H + ][OH - ] = 1.0  (3) (4) (5) We need to substitute to get things in terms of Ca 2+ or H + [H + ] + 2[Ca 2+ ] = [OH - ] + [F - ] 2[Ca 2+ ] = [F - ] + [HF]

15 K sp = [Ca 2+ ][F - ] 2 = 3.9  K w = [H + ][OH - ] = 1.0  [Ca 2+ ] = [F - ] + [HF] Combine eq. 2 and eq. 5, we have… (B)

Combine (B) and eq. 1, we have… (B) (C) It’s much simpler if we can consider the pH fixed (how would we do that?)

If we can fix pH… pH = 3.00 [H + ] = 1.0×10 -3 M KwKw [HF - ] =1.5[F - ] KbKb [F - ] =0.80[Ca 2+ ] Mass [Ca 2+ ] = 3.9×10 -4 M K sp [OH - ] = 1.0× M

18 The [Ca] removed from marble stone (largely dissolution of CaCO 3 ) increases as the [H + ] of acid rain increases. Applications of coupled equilibria in the modeling of environmental problems CaCO 3 (s) + 2H + (aq)  Ca 2+ (aq) + CO 2 (g) + H 2 O(l) SO 2 (g) + H 2 O(l)  H 2 SO 3 (aq) oxidation H 2 SO 4 (aq) Deposits include CaSO 4 2H 2 O (gypsum), which accumulates creating a black residue.

19 Total [Al] as a function of pH in 1000 Norwegian lakes. Acid rain also releases Al, Hg, and Pb into the environment.