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# lecture+11sf - Common Ion Effect Calculate[H3O percent...

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Common Ion Effect Calculate [H 3 O + ], percent ionization for: (a) 1.0 M HF (b) Mixture of 1.0 M HF and 1.0 M NaF K a = 7.2 x 10 -4 for HF Are the results consistent with LeChatelier’s Principle? (a) HF + H 2 O   H 3 O + + F - 1.0 - x x x x 0 . 1 x 2 - = 7.2 x 10 -4 Solving: x = [H 3 O + ] = 0.027 M pH = -log (0.027) = 1.57 percent ionization = x 100 = 2.7% (b) A mixture of 1.0 M HF and 1.0 M NaF initially has [HF] = 1.0 and [F - ] = 1.0, where the F - ions come from the complete dissociation of NaF into Na + and F - ions. HF + H 2 O H 3 O + + F - 1.0 - x x 1.0 + x = 7.2 x 10 -4 x = [H 3 O + ] = 7.2 x 10 -4 M pH = -log (7.2 x 10 -4 ) = 3.14 Percent ionization = 0 . 1 10 x 2 . 7 4 - x 100= 0.072% Note that 0.072% ionization (in the HF, NaF mixture) is much less than 2.7% ionization (in the 1.0 M HF alone). When we add extra F - (from NaF) the HF equilibrium gets shifted to the left, using up some of the extra F - (LeChatelier’s Principle). This is called the common ion effect , and we’ll

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see it again when we study solubility equilibrium. The “common ion” is F - . The extra F - from NaF shifts the HF equilibrium to the left, which lowers the percent ionization of HF. Our calculations are in agreement with LeChatelier’s Principle. Buffers The solution in the above example, in which HF and F - are both 1.0 M is called a buffer . A buffer contains both an acid together with its conjugate base, both present in appreciable concentration, and close (in the ballpark) to a 1:1 mol ratio. Compare three related, but different solutions, and calculate pH of each solution. K a = 1.8 x 10 -5 for acetic acid. (a) 0.50 M CH 3 COOH (acetic acid). (b) 0.50 M CH 3 COONa (sodium acetate) (c) Mixture of 0.50 M CH 3 COOH and 0.50 M CH 3 COONa Solution (a) is a weak acid. CH 3 COOH + H 2 O H 3 O + + CH 3 COO - 0.50 - x x x x 50 . 0 x 2 - = 1.8 x 10 -5 Solving: x = [H 3 O + ] = [CH 3 COO - ] = 0.0030 M pH = - log (0.0030) = 2.52 There is a small amount (0.0030 M) acetate ion present with the 0.50 M acetic acid. This is not a buffer. The ratio of acetic acid/acetate ion is very far from 1:1, and the acetate ion concentration is very small. Solution (b) is a weak base. The acetate ion is the conjugate base of acetic acid. For acetate ion:
K b = a w K K = 5 14 10 x 8 . 1 10 x 0 . 1 - - = 5.6 x 10 -10 CH 3 COO - + H 2 O CH 3 COOH + OH - 0.50 - x x x x 50 . 0 x 2 - = K b = 5.6 x 10 -10 x = [OH - ] = [CH 3 COOH] = 1.7 x 10 -5 pOH = -log (1.7 x 10 -5 ) = 4.78 pH = 14.00 - 4.78 = 9.22 There is a small amount (1.7 x 10 -5 M) acetic acid present with the 0.50 M acetate ion. This is not a buffer. The ratio of acetic acid/acetate ion is very far from 1:1, and the acetic acid concentration is very small. Solution (c) is a buffer, containing both HC

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## This note was uploaded on 03/08/2011 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.

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lecture+11sf - Common Ion Effect Calculate[H3O percent...

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