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How many grams of sodium acetate must be added to 100 mL of 0.150M acetic acid in order to produce a solution with a pH of 4.620? Assume that no volume change occurs.

Please check Answer:

Rearranging the H-H Eqn and solving for the ratio of base to acid:  

            pH - pKa = log([A-] / [HA])

            4.620 - 4.754 = log ([HC2H3O2] / [NaC2H3O2]

            -0.134 = log ([0.150 M] / [NaC2H3O2])

To remove the log function, take the antilog, or 10x of both sides:

            10^ -0.134 = [0.150] / [NaC2H3O2]

            0.735 = [0.150] / [NaC2H3O2]

Solve for the concentration of the ammonium chloride:

            [NaC2H3O2] = 0.150 / 0.735

            [NaC2H3O2] = 0.204 M

100 mL x (0.204 M NaC2H3O2) x (1 L/ 1000 mL) x (82.0343 g/mol) = 1.673 g NaC2H3O2

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