Chapter 9: POLYPROTIC ACID-BASE EQUILIBRIA

1 3/28/20141 Chapter 9: POLYPROTIC ACID-BASE EQUILIBRIAI mages available at acid , rhubarb and rhubarb pie1 Understanding POLYPROTIC SpeciesDiprotic acids, H2A vs. Dibasic species, A2-Triprotic acids, H3A vs. Tribasic species, A3-Ex. H2SO4, H2CO3Ex. SO42-, CO32-Ex. H3PO4, Ex. PO43-H3C6H5O7(Citric acid ) C6H5O73-2 Can donate 2 H+ Can accept 2 H+ Can donate 3 H+ Can accept 3 H+3/28/20142 Diprotic Acids and BasesGeneral formulas: H2A= fully acidic form HA-= intermediate form; amphoteric A2-= fully basic or fully deprotonated form3 EQUILIBRIA involved: Diprotic Acids and Bases -+223a1H A + H O HA + H OK First dissociation:Second dissociation:Diprotic acid , H2A Dibasic species, A2-2---2b1A + H O HA + OHK First hydrolysis:--22b2HA + H O H A + OHK Second hydrolysis:Q.

2 How do we calculate Kb1and Kb2from Kavalues?--2+23a2HA + H O A + H OK 43/28/20143 Note that H2A and HA-species in the Ka1expression both appear in the Kb2expression. Similarly, the conjugates HA-and A2-in the Ka2expression both appear in the ,Proof:2231aH A H OHAH OK + ++ 222bHAH OH A OHK ++ +232wH OH OOHK+ + Kw= Ka1 x Kb25 POLYPROTIC acids: Amino AcidsNOTE: - COOH group is much more acidic (higher Ka; first to dissociate) than the NH3+ AcidsExample: Leucine, H2L Stepwise dissociation:7 Fully protonated form = fully acidic, H2A+Fully dissociated form = fully basic, A- Start with the fully acidic form, H2A+= H2L+Dibasic Species8 Stepwise hydrolysisof leucine: Start with the fully basic form, A-= L-3/28/20145pH Calculations: Diprotic Acids and Bases Problem: Find the pH and concentrations of H2SO3, HSO3-and SO32-in each of the following solutions.

3 (a) M H2SO3(b) M NaHSO3, and(c) M Na2SO3 Note that for diprotic acids and bases, there are 3 species in solution ( 3 unknowns: H2A, HA-and A2-) so we need 3 independent equations to solve the Calculation: Diprotic Acids and Bases 2231aH A H OHAH OK + ++ 1. The fully acidic form, H2 AApproximation: In a solution of H2A (Ex. M H2SO3), the 2nddissociation is usually negligible that H2A behaves as amonoprotic acid . Also, [A2-] 0 of pH and [species]Equil: F-xx x21axKF x= 103/28/20146 Fully acidic form (H2A) SOH OHSOH OK + ++ Problem (a): Find the pH and [H2SO3], [HSO3-] and [SO32-] in a M H2SO3solution.

4 Ka1= x 10-2; Ka2= x 10-8 Equil: 10( )aKxx == xcannot be ignored since Ka1isn t too 10xx + Solve for x using quadratic 10[] []xMH OHSO + === M - [] SOM=23x[] 10 HSOM =23[] 0 SOM 11123/28/2014713pH Calculations: Diprotic systems OHAOHK ++ 2. The fully basic form, A2-Approximation: In a solution of A2-(Ex. M Na2SO3), the 2ndhydrolysis is usually negligible that A2-behaves as amonobasic , [H2A] 0 of pH and [species]Equil: F-yy y21byKF y= pOH = -log (y)pH = 14 - pOHRecall: Kb1= Kw/Ka2143/28/20148 Problem: Find the pH and concentrations of H2SO3, HSO3-and SO32-in each of the following solutions:(a) M H2SO3 - DONE!

5 (b) M NaHSO3, and (c) M Na2SO3Ka1= x 10-2; Ka2= x 10-8 Answer: pH = ; [H2SO3] 0 M; [SO32-] = M M and [HSO3-] = [OH-] = x 10-5M 15pH Calculations: Diprotic systems OH A OHKHA ++ 3. The intermediate (amphoteric) form, HA-Q. What is the predominant species in a solution of HA-?Compare Ka2and Kb2equilibria: HA-can act as an acid or a base2232aH OAH OKHA + ++ Dissociation:Hydrolysis: HA-will dissociate/hydrolyze to form A2-and H2 AApproximation: [HA-] FHA-= FNaHAor FKHA163/28/20149 The intermediate form, HA-(Cont.)

6 Calculation of pH and [species]pH = -log [H+]WhereK1= Ka1K2= Ka2F = FHA-Quick check: pH = (pK1+ pK2) Solve for [H2A] and [A2-] using [H+] above and K1& K2equilibria17 The intermediate form, HA-(Cont.) Solving for [H2A] and [A2-]:112[][[]]aHHAKH AK+ ==21[[]][]HHATh sAuHK+ =22[][],[]K HAAL ikewiseH +=2222:[][[]]aFrom nd dissociationHKKAHA+ ==183/28/201410 Problem: Find the pH and concentrations of H2SO3, HSO3-and SO32-in each of the following solutions:(a) M H2SO3 - DONE! (b) M NaHSO3, and(c) M Na2SO3 DONE!

7 Ka1= x 10-2; Ka2= x 10-8 Answer: pH = ; [HSO3-] M, [H2SO3] = x 10-4M; [SO32-] = x 10-4M 19 Triprotic Acids and BasesExample: H3PO4; PO43-3422431aH POH OH POH OK + ++ 2242432aH POH OHPOH OK + ++ 2342433aHPOH OPOH OK + ++ Successive dissociation:Successive hydrolysis:324241bPOH OHPOOHK ++ 242242bHPOH OH POOHK ++ 242343bH POH OH POOHK ++ 203/28/201411 Use handout on pH calculations involving triprotic systems21 Treatment of triprotic systems1. H3A is treated as monoprotic weak acid .

8 Ka1= H2A-is treated as the intermediate form of a diprotic HA2-is also treated as the intermediate form of a diproticacid. However, HA2-is surrounded by H2A-and A3-, so the equil. constants to use are Ka2(= K2) and Ka3(= K3)4. A3-is treated as monobasic. Kb1= Kw/Ka3223/28/201412 What is the major species at a given pH?1. When pH < pK1, H2A predominates2. When pH = pK1, [H2A] = [HA-]3. When pK1> pH < pK2, [HA-]predominates4. When pH = pK2, [HA-] = [A2-]5. When pH > pK2, [A2-]predominatesQ. Which of the species above predominate at pH pH pH Answer: A2-, HA-, H2A 2324