CEE March FIRST EXAM (Solutions)

Transcription

1 CEE March 015 FIRST EXAM (Solutions) Closed book, one page of notes allowed. Answer any 4 of the following 5 questions. Please state any additional assumptions you made, and show all work. You are welcome to use a graphical method of solution if it is appropriate. Miscellaneous Information: R = cal/mole K = J/mole K Absolute zero = C 1 joule = 0.39 calories 0 C = wicked cold 1. (5%) Use the graphical solution to determine the ph and complete solution composition for 1 liter of pure water to which you ve added 10.5 moles of NaH. Graph paper is attached to this exam for this purpose. PBE which reduces to: [H 3 ] + [H + ] = [OH - ] + [H ] + [ ] [H + ] = [H ] Equality (intersection) occurs at ph ~ 5.35 Species pc C (Molar) H x 10-6 OH x 10-9 H x H x 10-4 H x x Na x 10-4

2 H + OH - - H 3 H H PBE Solution -6 Log C ph. (5%) Determine the ph and solution composition of a mixture of 10.5 moles of NaH plus 10 moles of Sodium Acetate in 1 liter of water. Please use a graphical solution for this one too. PBE which reduces to: [H ] + [ ] + [OH - ] = [H + ] + [H 3 ] + [HAc] [H ] = [HAc] Equality (intersection) occurs at ph ~ 6.

3 Species pc C (Molar) H x 10-7 OH x 10-8 H x H x 10-4 H x x HAc x 10-5 Ac x 10 Na x 10 Log C vs ph Diagram H+ HAc Ac - OH - H - H PO HPO 3 PO PBE Solution -5-6 Log C ph

4 3. (5%) Determine the complete solution composition of a solution of moles of Iodic Acid (HIO 3 ) in 1 Liter of water, but for this one use an algebraic solution. Do this again, but this time assume you re adding only moles of the conjugate base [i.e., Sodium Iodate (NaIO 3 )] in 1 Liter of water. Remember to make simplifying assumptions. A. Make acidic solution assumption, which leads to: [ H + K ] + K a a + [H + ] = x 10-5 ph = K C but easier to make the combined acidic solution and strong acid assumption, which leads to: [ H + ] C [H + ] = x 10-5 ph = a B. Make basic solution assumption, which leads to: [ OH ] K K b + b + 4K C b [OH - ] = x 10-9 [H + ] = x 10-6 ph = 5.15 This is not a basic so the assumption is not good. Instead try the weak base assumption (i.e., [IO 3 - ] >> [HIO 3 ]; since the strong acid assumption worked for the conjugate acid), which leads to: [ OH ] K C + b K w

5 [OH - ] = x 10-7 [H + ] = x 10-8 ph = now check the assumption that [IO 3 - ] >> [HIO 3 ] at this ph ( ). Ka = [H + ][IO 3 - ]/[HIO 3 ] =10-7 [IO 3 - ]/[HIO 3 ] [IO 3 - ]/[HIO 3 ] = Yes, the assumption is good 4. (5%) Repeat problem #1, but this time add 0.1 M of NaCl as well as the 10.5 moles of NaH to your liter of water The ph shift pertains essentially to the second pk only. At I=0, pk =7.. so, now calculate its value at I = { H }{ HA } [ H ] γ + [ HA ] γ { H }[ HA ] γ H HA HA K = = = H A H A γ H A γ and = { } + { H }[ A ] γ HA [ ] HA γ H A K ' [ ] + { H }[ A ] K = γ H H A = K γ [ ] [ ] HA γ γ HA HA A H A H A Use the Guntelberg approximation: log f = 0.5z I 1+ I I = 0.1 use Guntelberg Phosphoric acid pk = 7. at I=0 log(fha)= fha = log(fha-)= fha- = fh+ = pk' = at I = 0.1

6 Everything else just follows what was done in problem #1: PBE which reduces to: [H 3 ] + [H + ] = [OH - ] + [H ] + [ ] [H + ] = [H ] At this point you could recognize that the intersection will shift half a ph unit for each ph unit shift in the pk. So for a 0.36 ph unit drop in the pk, the intersection should drop 0.18 ph units or from about 5.35 to You could also re-draw the original figure and it would look like this: Log C vs ph Diagram H + OH - - H 3 H H PBE Solution -6 Log C ph

7 Equality (intersection) occurs at ph ~ 5.15 You can them read off the concentrations as before. Strictly speaking, to get the concentrations of the two extreme phosphate species (H 3 and ) you d need to do a similar calculation for the pk 1 and the pk 3 and re-graph. Also, the exact PBE would require that you calculate a new H+ line as the ideal relationship between the molar [H + ] and ph (which is related to activity, i.e., ph={h + }) is now affected by ionic strength too.

8 5. (5%) True/False. Mark each one of the following statements with either a T or an F a. T The ph of pure water to which you add 10 M NaH and 10 Na H is about equal to the second pka for the phosphate system (i.e., ph ~ 7.) b. F Conditional equilibrium constants are independent of ionic strength c. F The sum of any pka and the pkb of its conjugate base is always 1. d. T ph = pka at the mid-point of a titration e. F Nitric acid always completely donates its proton to the solvent it is dissolve in, regardless of the nature of that solvent f. T The principle of electroneutrality is always observed in aqueous solutions g. F Non-carbonate hardness is equal to the magnesium concentration h. F Increases in ionic strength have no effect on species with zero charge. i. F Nitrate is a very strong base in water j. F The value of αo plus α 1 must always equal 1 for any diprotic acid system. Note the original had a typo in question a, so I accepted both T and F for this one.

9 Selected Acidity Constants (Aqueous Solution, 5 C, I = 0) NAME FORMULA pka Perchloric acid HClO4 = H + + ClO STRONG Hydrochloric acid HCl = H + + Cl - Sulfuric acid HSO4= H + + HSO4 - (&) ACIDS Nitric acid HNO3 = H + + NO Hydronium ion H3O + = H + + HO 0 Trichloroacetic acid CCl3COOH = H + + CCl3COO Iodic acid HIO3 = H + + IO3-0.8 Bisulfate ion HSO4 - = H + + SO4 Phosphoric acid H3PO4 = H + + HPO (&7.,1.3) o-phthalic acid C6H4(COOH) = H + + C6H4(COOH)COO-.89 (&5.51) Citric acid C3H5O(COOH)3= H + + C3H5O(COOH)COO (&4.77,6.4) Hydrofluoric acid HF = H + + F - 3. Aspartic acid CH6N(COOH)= H + + CH6N(COOH)COO (&9.8) m-hydroxybenzoic acid C6H4(OH)COOH = H + + C6H4(OH)COO (&9.9) p-hydroxybenzoic acid C6H4(OH)COOH = H + + C6H4(OH)COO (&9.3) Nitrous acid HNO = H + + NO Acetic acid CH3COOH = H + + CH3COO Propionic acid CH5COOH = H + + CH5COO Carbonic acid HCO3 = H + + HCO (&10.33) Hydrogen sulfide HS = H + + HS (&13.9) Dihydrogen phosphate HPO4 - = H + + HPO4 7. Hypochlorous acid HOCl = H + + OCl Boric acid B(OH)3 + HO = H + + B(OH)4-9. (&1.7,13.8) Ammonium ion NH4 + = H + + NH3 9.4 Hydrocyanic acid HCN = H + + CN p-hydroxybenzoic acid C6H4(OH)COO - = H + + C6H4(O)COO 9.3 Phenol C6H5OH = H + + C6H5O m-hydroxybenzoic acid C6H4(OH)COO - = H + + C6H4(O)COO 9.9 Bicarbonate ion HCO3 - = H + + CO Monohydrogen HPO4 = H PO4 phosphate Bisulfide ion HS - = H + + S 13.9 Water HO = H + + OH Ammonia NH3 = H + + NH - 3 Methane CH4 = H + + CH3-34

10 Species o H f o G f kcal/mole kcal/mole Ca + (aq) CaCO 3 (s), calcite CaO (s) C(s), graphite 0 0 CO (g) CO (aq) CH 4 (g) H CO 3 (aq) HCO - 3 (aq) CO 3 (aq) CH 3 COOH CH 3 COO -, acetate H + (aq) 0 0 H (g) 0 0 HF (aq) F - (aq) Fe + (aq) Fe +3 (aq) Fe(OH) 3 (s) NO - 3 (aq) NH 3 (g) NH 3 (aq) NH + 4 (aq) HNO 3 (aq) O (aq) O (g) 0 0 OH - (aq) H O (g) H O (l) PO 4 (aq) HPO 4 (aq) H PO - 4 (aq) H 3 (aq) SO HS - (aq) H S(g) H S(aq)

11 Log C T ph

12 Log C T ph