1. A 0.0450 M solution of benzoic acid has a pH of 2.71. Calculate pKa for this

Question

1. A 0.0450 M solution of benzoic acid has a pH of 2.71. Calculate pKa for this acid.

2. A 0.0460 M solution of HA is 0.80% dissociated. Calculate pKa for this acid.

3. Consider a reaction mixture containing 100.0 mL of 0.112 M borate buffer at pH = pKa = 9.24. At pH = pKa, we know that [H3BO3] = [H2BO3] = 0.0560 M. Suppose that a chemical reaction whose pH we wish to control will be generating acid. To avoid changing the pH very much we do not want to generate more acid than would use up half of the [H2BO3].

a ) How many moles of acid could be generated without using up more than half of the [H2BO3]?

b) What would be the pH?

Explanation / Answer

1. Lets consider HA as bezoic acid then HA => H+ + A-

As we know pH = -log[H+] therefore H+ = 10^-pH = 10^-2.71 = 0.00194

Also as conc of H+ = Conc. of A-

therefore Conc of H+ and A- = 0.00194

where as conc of HA is 0.0450M - x

Write the equilibrium reaction HA => H+ + A-

From Above Ka = [H+][A-]/[HA] = (0.00194)^2/0.045-0.00194 = 3.7X10^-6/0.04306 = 8.74X10^-5

pKa = -log[Ka] = -log [8.74X10^-5] = 4.058

2. The amount of A- and H+ produced is 80% of the 0.0460M starting value.

80% of 0.046M = 0.0368

From which we can assuem that this much HA is lost, so at equilibrium conc of HA is 0.046 - 0.0368 = 0.0092

Ka = [H+][A-] / [HA]

Ka = [0.0368[0.0368] / [0.0092] = 0.1472

pKa = -log[Ka] = -log[0.1472] = 0.83

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