Question

(6) Let's see how the spectator" ions present in solution affect the pH of a weak acid (a) What is the pH of a 0.0200 M benzoic acid (Ka = 6.28 x 10) solution in water? (b) What is the % dissociation of the benzoic acid in pure water? Now we'll add some "spectator" ions to the solution. If the benzoic acid is dissolved in 0.10 M CaCl2 instead of pure water: (c) Calculate the ionic strength of the solution (assume that the contribution from the benzoic acid is negligible). (d) Find the appropriate hydrated radii (a) and calculate the activity coefficients (Y) for both the H and the benzoate ion.

Answer 1

Part (a).

pH = 1/2 (pKa - Log[benzoic acid])

= 1/2 {-Log(6.28*10^-5) - Log(0.02)}

= 1/2 (4.2 + 1.7)

Therefore, pH = 2.95

Part (b).

The % dissociation of benzoic acid in pure water ($\alpha$) = (Ka/[benzoic acid])^1/2 * 100 = (6.28*10^-5/0.02)^1/2 *100 = 0.056*100 = 5.6%