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Aqueous Equilibria, Part Two AP Chemistry. The Common-Ion Effect Consider a mixture of a “weak” and a “common-ion,” soluble salt. CH 3 COOH CH 3 COO –

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Presentation on theme: "Aqueous Equilibria, Part Two AP Chemistry. The Common-Ion Effect Consider a mixture of a “weak” and a “common-ion,” soluble salt. CH 3 COOH CH 3 COO –"— Presentation transcript:

1 Aqueous Equilibria, Part Two AP Chemistry

2 The Common-Ion Effect Consider a mixture of a “weak” and a “common-ion,” soluble salt. CH 3 COOH CH 3 COO – + H + Since NaCH 3 COO is strong, adding it to the solution… By Le Chatelier… The result is that… CH 3 COOHNaCH 3 COO greatly [CH 3 COO – ]. shift [H + ] and pH sodium acetate

3 This illustrates the common-ion effect: “The dissociation of a ‘weak’ DECREASES when a strong, common-ion salt is added to the solution.” For 100 HBrOs in soln… HBrOH + + BrO – 9010 If we add a “strong,” BrO – salt (like KBrO), the amount of BrO – goes up, and the amount of HBrO that dissociates goes down (by Le Chatelier).

4 Find the pH of a solution containing 0.085 M nitrous acid (K a = 4.5 x 10 –4 ) and 0.10 M potassium nitrite. HNO 2 H + + NO 2 – 0.085 – x 0 + x 0.10 + x = 0.10x + x 2 0.085 – x K a = 4.5 x 10 –4 So… x = [H + ] = 3.825 x 10 –4 MpH = 3.42 X X Find the pH of 0.085 M nitrous acid, on its own. 4.5 x 10 –4 = x2x2 0.085 – x X So… x = [H + ] =6.185 x 10 –3 MpH = 2.21 HNO 2 DID dissociate less when the common-ion salt was present.

5 Buffered Solutions (“buffers”) -- a mixture of… -- e.g., a “weak” (either acid or base) and a common-ion salt HOAc and NaOAc solns based on the common-ion effect HF and KF NH 3 and NH 4 Cl HBrO 2 and Ca(BrO 2 ) 2 buffer capacity: the amount of acid or base the buffer can “neutralize” before the pH begins to change appreciably -- buffer capacity increases with… greater amounts of acid/base AND common-ion salt -- they resist  pH

6 “You betcha.” Find the pH of a buffer that is 0.12 M lactic acid, HC 3 H 5 O 3 (K a = 1.4 x 10 –4 ) and 0.10 M sodium lactate. (NaC 3 H 5 O 3 ) HC 3 H 5 O 3 H + + C 3 H 5 O 3 – 0.12 – x 0 + x 0.10 + x “Shortcut OK?” 1.4 x 10 –4 = 0.10x + x 2 0.12 – x x = [H + ] = 1.68 x 10 –4 MpH = 3.77 X X


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