255 grams of barium acetate BaC2H3O22 is added to 5000 ml of

25.5 grams of barium acetate, Ba(C2H3O2)2 is added to 500.0 ml of 0.100 M sodium acetate, NaC2H3O2. Calculate the pH of the resulting solution. [You may assume no change in volume upon the addition of the solid.] HC2H3O2

Ka = 1.8×10–5

Ba(C2H3O2)2 = 255 g/mol

NaC2H3O2 = 82.0 g/mol

Solution

molarity of [BaC2H3O2] = 25.5 g/255 g/mol x 0.5 L = 0.2 M

molarity [NaC2H3O2] = 0.1 M

total molarity of [C2H3O2-] = 2 x 0.2 M x 0.1 M = 0.5 M

C2H3O2- + H2O <==> HC2H3O2 + OH-

let x amount hydrolyzed

Kb = Kw/Ka = [HC2H3O2][OH-]/[C2H3O2-]

1 x 10^-14/1.8 x 10^-5 = x2/0.5

x = [OH-] = 1.67 x 10^-5 M

pOh = -log[OH-] = 4.78

pH = 14 - pOH = 9.22