Ch 7. Systematic Treatment of Equilibrium and Activity

Transcription

1 Ch 7 Systematic Treatment of Equilibrium and Activity

2 Systematic Treatment of Equilibrium Complicated systems have several unknown variables species of unknown concentration Generally, you need to come up with as many independent equations as unknown variables Strategy (ie. PbI 2 problem) Write down all pertinent equilibrium expressions Write mass balance equations Write down charge balance equations Substitution to reduce problem to one equation with one variable

3 Mass Balance Let s look at a triprotic system, such as H 3 PO 4, as an example There are four different species within this system H 3 PO 4 H 2 PO - 4 HPO 4 2- PO 4 3-

4 0.050 M NaH 2 PO = [H 3 PO 4 ] + [H 2 PO 4- ] + [HPO 4 2- ] + [PO 4 3- ]

5 Charge Balance All solutions must be electrically neutral!!!!!!!! Which means they carry a no net charge. mole positive charges = moles negative charges ex. A dilute solution of acetic acid, HA What species exist in solution? HA, A -, OH -, H 3 O + From HA H + + A - and 2H 2 O H 3 O + + OH -

6 A mole of A - carries a mole of negative charge. A mole of OH - carries a mole of negative charge. A mole of HA carries no charge. A mole of H 3 O + carries a mole of positive charge. Therefore, the charge balance equation is [H 3 O + ] = [A - ] + [OH - ]

7 Charge Balance for the M NaH 2 PO 4 solution 1 mole of H 3 PO 4 carries no moles of charge 1 mole of H 2 PO 4- carries 1 mole of (-) charge 1 mole of HPO 2-4 carries 2 moles of (-) charge 1 mole of PO 3-4 carries 3 moles of (-) charge Therefore, [Na + ] + [H 3 O + ] = [OH - ] + [H 2 PO 4- ] + 2[HPO 4 2- ] + 3[PO 4 3- ]

8 Why is this useful? In many instances when we are dealing with very complex equilibria, it is useful to use mass balance and charge balance in efforts to reduce an expression to one variable. Stay tuned. We will utilize mass and charge balance in the next couple of chapters

9 Write a mass balance and charge balance equation for a M solution of H 2 SO 4 (HSO 4- is a very strong weak acid)

10 Mass Balance M = [H 2 SO 4 ]+[HSO 4- ]+[SO 2-4 ] Charge Balance [H 3 O + ] = [HSO 4- ] + 2[SO 2-4 ] + OH -

11 Chapter 8 Activity

12 CH 313 Experiment Solubility of CaSO 4 solution Sat. CaSO 4 in H 2 0 [Ca 2+ ] 1.5E-03 M 8.8 E-04 M Sat. CaSO 4 in Na 2 SO 4 Sat. CaSO 4 in KNO 3 2.3E-03 M

13 The true nature of ionic species in solution Ions are charged molecules As a result, they tend to attract polar solvent molecules (like water, for instance) and other ions Hydrated radius Ionic atmosphere The thickness of the ionic atmosphere is a function of the ionic strength of the solution the concentration of charge

14 Hydrated radius _ + _ + water Chloride Sodium ion

15 Ionic Atmosphere and Shielding Low ionic strength High ionic strength _ + - Calcium ion Sulfate ion Sodium ion Chloride ion

16 Activity All ionic species in any equilibrium expression are more accurately expressed as activities. A Ca2+ = [Ca 2+ ]γ Ca2+ or A + H 2 O A - + H 3 O + Ka = A A- A H3O+ / [A] = [A - ]γ A- [H 3 O + ]γ H3O+ / [A] where γ is the activity coefficient and is a function of the ionic radius of the ion and the ionic strength of the solution.

17 Activity - continued The higher the ionic strength of the solution, the larger the smaller the activity coefficient. Rationalization: At higher ionic strengths the ion cloud around any ion is thicker, which weakens the attractive forces between the ion and its counterpart, inhibiting recombination.

18 Ionic strength A measure of the concentration of ions in solution μ = ½ c i z i M Ca(NO 3 ) 2 μ = ½ ([NO 3- ](-1) 2 + [Ca 2+ ](2+) 2 ) = ½ {(0.02)(-1) 2 + (0.01)(2) 2 } = 0.03 M γ Ca2+@μ=0.03 =? Use extended Debye-Huckle eq or Table 8.1 and extrapolation

19 Take home message At high ionic strengths, solubility increases slightly (by a factor of 2-10). ph is influenced by ionic strength Weak acid and base dissociation is influenced a little by ionic strength Significant figures?

20 Example 8-12 Solubility of Hg 2 Br 2 in pure water in M KNO 3 in M KNO 3 in M KNO 3

21 In pure water Hg 2 Br 2 Hg Br - Ksp = [Hg 2+ 2 ] γhg22+ [Br - ] 2 γ 2 Br- 2[Hg 2+ 2 ] = [Br - ] and let [Hg 2+ 2 ] = x Ionic strength is very low. γhg22+ and γ Br- are close to 1.00 so, Ksp = 4 [x] 3 = x = M

22 Activity coefficient as a function of ionic strength Table 8-1 μ = μ = 0.01 μ = 0.1 Hg Br

23 in M KNO3 μ = ½ ((.001)(+1) 2 + (0.001)(-1) 2 = M γ μ = = γ 2 μ = = Ksp = 4 γ Hg22+ γ Br-2 [x] 3 = x = M

24 in M KNO3 μ = ½ ((.01)(+1) 2 + (0.01)(-1) 2 = 0.01 M γ μ = = γ 2 μ = = Ksp = 4 γ Hg22+ γ Br-2 [x] 3 = 5.6E-23 x = M

25 in M KNO3 μ = ½ ((0.1)(+1) 2 + (0.1)(-1) 2 = 0.1 M γ μ = = γ 2 μ = = 0.75 Ksp = 4 γ Hg22+ γ Br-2 [x] 3 = 5.6E-23 x = M

26 solubility vs log(μ) solubility of Hg2Br2 4.7E E E E E E log(μ)

27 ph and ionic strength True definition of ph ph = -loga H+ = -log {[H + ]γ H+ } ph of a M HCl solution Ionic strength, μ, = 0.001; γ H+ = ph = -log(.001*.967) = 3.01 ph of a M HCl solution Ionic strength, μ, = 0.1; γ H+ = 0.83 ph = -log(0.1*0.83) = 1.08 ph of a M HCl/0.100 M NaCl solution Ionic strength, μ, = 0.1; γ H+ = 0.83 ph = -log(0.001*0.83) = 3.08

28 What is the concentration H + and OH - in a M NaCl solution? Kw = [OH - ]γ OH- [H + ]γ H+ = At μ = 0.100, γ OH- = 0.76 and γ H+ = 0.83 H 2 O is the only source of H + and OH - so let x = [H + ] = [OH - ] Kw = (0.76)(0.83) x 2 x = M