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