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Equilibria:Polyprotic weak acids For example, phosphoric acid, H 3 PO 4 H 3 PO 4  H 2 PO 4 -1 + H +1 K A1 = 7.5*10 -3 H 2 PO 4 -1  HPO 4 -2 + H +1 K.

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Presentation on theme: "Equilibria:Polyprotic weak acids For example, phosphoric acid, H 3 PO 4 H 3 PO 4  H 2 PO 4 -1 + H +1 K A1 = 7.5*10 -3 H 2 PO 4 -1  HPO 4 -2 + H +1 K."— Presentation transcript:

1 Equilibria:Polyprotic weak acids For example, phosphoric acid, H 3 PO 4 H 3 PO 4  H 2 PO 4 -1 + H +1 K A1 = 7.5*10 -3 H 2 PO 4 -1  HPO 4 -2 + H +1 K A2 = 6.2*10 -8 HPO 4 -2  PO 4 -3 + H +1 K A3 = 4.8*10 -13 H +1 T = H +1 1 + H +1 2 + H +1 3 ~ H +1 1

2 Equilibria: Salts Illustrated using a simple =1/-1 salt There are 4 reactions to consider. 1.CatAn  Cat +1 + An -1 2.Cat +1 + H 2 O  CatOH + H +1 K A = K W /K B 3.An -1 + H 2 O  Han + OH -1 K B = K W /K A 4.H 2 O  H +1 + OH -1 K W

3 Equilibria: Salts We may classify salts into 4 groups, depending on the source of the cation and anion. SBSA (NaCl), WBSA (NH 4 Cl), SBWA (NaF), WB WA (NH 4 F). The pH of the final solution will be influenced by one or more of the previous reactions. Reactions influencing solution pH Rxn #SBSAWBSASBWAWBWA 2XX 3XX 4XXXX

4 Equilibria: Salts WAWB salts are especially interesting. The pH is influenced by both the cation and anion hydrolysis reactions. 1.Cat +1 + H 2 O  CatOH + H +1 K A = K W /K B 2.An -1 + H 2 O  Han + OH -1 K B = K W /K A The pH of the final solution will depend on the size of K A & K B. K A > K B, acidic, etc.

5 Equilibria: Titrations If one titrates an acetic acid solution with a sodium hydroxide solution. There are 4 stages in the titration. 1.Only acetic acid, pH = -log(C A K A )/2 2.Buffer, pH = pK A + log [A-] [HA] 3.WASB salt, pOH = -log(C A K B )/2 note, must find new “acid” concentration 3.Strong base pH = 14 - pOH

6 Equilibria: Solubility & Solubility Products If a salt is “insoluble” or sparingly soluble, one may write an equilibrium expression for the reaction with water. M X A Y(S)  XM +Y (aq) + YA -X (aq) K SP = [M +Y ] X [A -X ] Y e.g. CaF 2(S)  Ca +2 (aq) + 2F -1 (aq) K SP = [Ca +2 ] 1 [F -1 ] 2 K SP = 4.2*10 -11

7 Equilibria: Solubility & Solubility Products We prepare a solution which is 1.0*10 -4 M in copper(I) and 2.0*10 -3 M in lead(II). To this solution we slowly add sodium iodide. Which will precipitate first, CuI or PbI 2 ? The K SP ’s are respectively, 5.3*10 -12 and 1.4*10 -8.


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