Consider a reaction mixture containing 100.0 mL of 0.100M borate buffer at pH =

Question

Consider a reaction mixture containing 100.0 mL of 0.100M borate buffer at pH = pka = 9.24. At ph = pka, we know that [H3BO3] = [H2BO3-] = 0.050 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-]. How many moles of acid could be generated without using up more than half [H2BO3-]? What could be the pH?

This is question #28 of Chapter 9 (Quantitative Chemical Analysis, Ninth Edition, Daniel C. Harris)

Explanation / Answer

In 100 mL buffer you have

0.050 mol/L x 0.100 L = 0.005 mol H2BO3-

The reaction may generate up to

0.005 mol /2 = 0.0025 mol acid (as H+).

If this quanty of acid is generated, the actual buffer will be

[H3BO3] = 0.050 mol/L + 0.025 = 0.075 M (50% increased)

[H2BO3-] = 0.050 mol/L - 0.025 = 0.025 M   (50% decreased)

pH = pKa + log ([H2BO3-]/[ H3BO3]) =

       = 9.24 + log (0.075/0.025) = 9.72

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