1 / 27

28-3-2011

28-3-2011. Diprotic and Polyprotic Acids. Have more than one acidic proton. If the difference between the K a for the first dissociation and subsequent K a values is 10 3 or more, the pH generally depends only on the first dissociation. Species concentrations of diprotic acids.

silas
Download Presentation

28-3-2011

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 28-3-2011

  2. Diprotic and Polyprotic Acids Have more than one acidic proton. If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.

  3. Species concentrations of diprotic acids Evaluate concentrations of species in a 0.10 M H2SO4 solution. Solution: H2SO4g H+ + HSO4– completely ionized (0.1–0.1) 0.10+y 0.10-y HSO4–D H+ + SO42–Ka2 = 0.012 0.10–y 0.10+y y Assume y = [SO42–] (0.10+y) y ————— = 0.012 (0.10-y) [SO42–] = y = 0.01M [H+] = 0.10 + 0.01 = 0.11 M; [HSO4–] = 0.10-0.01 = 0.09 M Y2 + 0.112 y – 0.0012 = 0 - 0.112+0.1122 + + 4*0.0012y = —————————————— = 0.0098 2

  4. Y2 + 0.112 y – 0.0012 = 0 - 0.112+0.1122 + + 4*0.0012y = —————————————— = 0.0098 2 Another approach: Two steps method Solution: H2SO4g H+ + HSO4–completely ionized (0.1–0.1) 0.10 0.10 In the 2nd equilibrium, substitute in [H+] = 0.1 M and [HSO4-] = 0.1 M HSO4–D H+ + SO42–Ka2 = 0.012 0.10–x 0.10+x x (0.10+x) x————— = 0.012, assume 0.1>>x, RE = {0.012/0.1}*100 = 12% (0.10-x) Assumption is not OK [SO42–] = y = 0.01M [H+] = 0.10 + 0.01 = 0.11 M; [HSO4–] = 0.10-0.01 = 0.09 M

  5. Species concentrations of weak diprotic acids Evaluate concentrations of species in a 0.10 M H2S solution. Solution: H2S D H+ + HS–Ka1 = 1.02*10-7 (0.10–x) x+y x-y Assume x = [HS–] HS–D H+ + S2–Ka2 = 1.0*10-13 x–y x+y y Assume y = [S2–] (x+y) (x-y) (x+y) y ————— = 1.02*10-7 ———— = 1.0*10-13 (0.10-x) (x-y) [H2S] = 0.10 – x = 0.10 M[HS–] = [H+] = x  y = 1.0*10–4 M; [S2–] = y = 1.0*10-13 M 0.1>> x >> y: x+ y = x-y = xx = 0.1*1.02*10-7 = 1.00*10-4y = 1*10-13

  6. Another approach: Two steps method Solution: H2S D H+ + HS–Ka1 = 1.02*10-7HS–D H+ + S2–Ka2 = 1.0*10-13 Ka1/Ka2 >102 , therefore the 2nd equilibrium can be neglected for the moment. From the 1st equilibrium: H2S D H+ + HS–Ka1 = 1.02*10-7 0.1 – x x x 1.02*10-7 = x2/(0.1-x), x = 10-4 and RE = {10-4/0.1}*100 = 0.1% (OK) Therefore, [H+] = [HS-] = 10-4M

  7. From the second equilibrium, we have: HS–D H+ + S2–Ka2 = 1.0*10-13 10-4 -x 10-4 +x x 1.0*10-13 = (10-4 +x)x/(10-4 -x) Assume 10-4 >>x X = 1.0*10-13 which is very small compared to 10-4 Therefore, the concentrations of the different species are: [H2S] = 0.10 – x = 0.10 M[HS–] = [H+] = 10-4 - x = 10–4 M; [S2–] = x = 1.0*10-13 M

  8. NH3(aq) + H2O (l) NH4+(aq) + OH-(aq) [NH4+][OH-] Kb = [NH3] weak base strength Kb Weak Bases and Base Ionization Constants Kb is the base ionization constant Solve weak base problems like weak acids except solve for [OH-] instead of [H+].

  9. HA (aq) H+(aq) + A-(aq) A-(aq) + H2O (l) OH-(aq) + HA (aq) H2O (l) H+(aq) + OH-(aq) Kw Kw Ka= Kb= Kb Ka Ionization Constants of Conjugate Acid-Base Pairs Ka Kb Kw KaKb = Kw Weak Acid and Its Conjugate Base

  10. KaKb and Kw A- + H2O DHA + OH- [HA] [OH-] [H+]Kb = ————— ——— [A-] [H+] [HA] = ———— [OH-][H+][A-][H+] 1 = —— KwKa H+ + BaseDConjugate_acid of Base+AcidD H+ + Conjugate_base of Acid- For example: NH3 + H2O DNH4+ + OH-Ka for NH4+ = Kw / Kb for NH3 HAD H+ + A-Kb for A- = Kw / Ka for HA Thus, KaKb = Kw for conjugate pairs.

  11. KaKb and Kw – another way HA + H2O (l) DH3O+ (aq) + A–(aq) KaofHA +) A– + H2O (l) DOH–(aq) + HA (aq) KbofA– 2 H2O (l) DH3O+ (aq) + OH– (aq) Kw = KaKb When you add two equations to get a third, what are the relationship between the K’s?

  12. Weak Bases Bases react with water to produce hydroxide ion.

  13. NH3(aq) + H2O(l) NH4+(aq) + OH−(aq) [NH4+] [OH−] [NH3] = 1.8  10−5 Kb = pH of Basic Solutions What is the pH of a 0.15 M solution of NH3?

  14. NH3(aq) + H2O(l) DNH4+ (aq) + OH−(aq) Initial 0.150.000.00 Change -x+x+x Equilibrium 0.15 – xxx 1.8*10-5 = x2/(0.15 – x) Kb seems very small, therefore assume that 0.15>>x X = 1.6*10-3 Relative error = {1.6*10-3/0.15}*100% = 1.1%, which is acceptable. [OH-] = 1.6*10-3 M, pOH = 2.79, therefore pH = 14-2.79 = 11.2

  15. Aniline is an organic chemical used in dye synthesis and is a weak base, with a Kb of 3.8 x 10-10. What is the pH of a 1.5 M ArNH2 solution? ArNH2 (aq) + H2O(l) ArNH3+(aq) + OH-(aq) [ArNH3+] [OH-] Kb = [ArNH2] Concentration (M) ArNH2 H2O ArNH3+ OH- Initial 1.5 ---- 0 0 Change -x ---- +x +x Equilibrium 1.5 - x ---- x x making the assumption: since Kb is small: 1.5 M - x = 1.5 M

  16. Substituting into the Kb expression and solving for x: [ArNH3+][OH-] (x)(x) Kb = = = 3.8 x 10-10 [ArNH2] 1.5 - x x = 2.4 x 10-5 , RE = {2.4*10-5/1.5}*100% = 1.6*10-3% x = 2.4 x 10-5 = [OH-] = [ArNH3+] Calculating pH: Kw 1.0 x 10-14 [H3O+] = = = 4.2 x 10-10 2.4 x 10-5 [OH-] pH = -log[H3O+] = - log (4.2 x 10-10) = 9.4 Or calculate pOH = 4.6 pH = 14 – 4.6 = 9.4

More Related