Prob Set _1 MCB 102 F13 ANS KEY

# The equilibrium constant for dissociation of an acid

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Unformatted text preview: he equilibrium constant for dissociation of an acid is: + - [H ] [A ] Ka = , with concentrations given in M. [HA] - We are given that the total concentration of the acid is 8.4 mM = 0.0084 M = [HA] + [A ] X2 Hence, if we set [A ] as X, Ka = 0.0084 M - X + + Since we are given the pH, and pH is defined as - log [H ], we can calculate [H ], as follows: + [H ] = 10-3.15 = 7.079 x 10-4 M + - - Furthermore, [H ] = [A ] and we set [A ] = X, thus [7.079 x 10-4 M] [7.079 x 10-4 M] Ka = -3 -4 = 6.52 x 10-5 M 8.4 x 10 M — 7.079 x 10 M Finally, the definition of pKa is - log Ka, hence this organic acid has a pKa = 4.18 Benzoic acid has a pKa = 4.18 Question 3. Citric acid (see right) is often used as a biological buffer for several reasons. As thoroughly as possible, answer each of the following queries about citrate below. (a) Citric acid and its salts are very soluble in water. For example, at 20˚ C, 100 ml of water will dissolve 146 g of citric acid. Why is citrate such a water-soluble molecule? Citrate is a very polar molecule. It has three potential charged groups (three -COO-) that will become hydrated by shells of water molecules, and it also has a hydroxyl group (-OH), which can form H-bonds with water molecules. For these reasons, citrate will be very well solvated in water and, hence, is highly soluble in aqueous solution. (b) Citrate is a symmetrical molecule with three chemically identical ionizable groups; yet, those ionizable groups have distinct pKa values (pK1 = 3.1, pK2 = 4.8, a...
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## This note was uploaded on 03/15/2014 for the course MCB 102 taught by Professor Staff during the Fall '08 term at Berkeley.

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