What mass (in grams) of potassium acetate, KC2H3O2, must be added to 124 mL of 0.010 M acetic acid to form a buffer solution with a pH of 4.5?

Question

 I cannot figure out what formula or how to start out the problem.

Please help.

Answer 1

The following equations apply here:  Let x be the grams of potassium acetate needed.

1)  K_a = 1.75 x 10^-5 = [H⁺][C₂H₃O₂^-] / [HC₂H₃O₂] = [H⁺] ( mol HC₂H₃O₂^-) / ( mol HC₂H₃O₂)

2)  [OH^-] = 1.0 x 10^-14 / [H⁺]                         (K_w expression)

3)  mol K⁺ + mol H⁺ = mol OH^- + mol C₂H₃O₂^-               (charge balance)

4)  mol HC₂H₃O₂ +  mol C₂H₃O₂^- = .124 L (.010 M) + (x / 98.1423 g/mol) = .00124 + x / 98.1423        (mass balance)

5)  [H⁺] = 10^-pH = 10^-4.5 = 3.1623 x 10^-5 M                 (pH = 4.50, given)

 

Now solve equation 1 for mol HC₂H₃O₂:  mol HC₂H₃O₂ = (3.1623 x 10^-5)(mol C₂H₃O₂^-) / 1.75 x 10^-5 = 1.807 (mol C₂H₃O₂^-)

Substitute this expression into equation 4:  (1.807 + 1) mol C₂H₃O₂^- = .00124 + x / 98.1423

Solve for mol C₂H₃O₂^- :  mol C₂H₃O₂^- = 4.4175 x 10^-4 + x / 275.485

Solve equation 3 for mol K⁺:  mol K⁺ = x / 98.1423 = -3.1623 x 10^-5 + 1.0 x 10^-14 / 3.1623 x 10^-5 + 4.4175 x 10^-4 + x / 275.485

combine x terms:  x (1/98.1423 - 1/275.485) = 4.0988 x 10^-4  -->  x = 0.06249 grams KC₂H₃O₂  (= 6.36 x 10^-4 mol)