Question

Calculate the pH of a solution of 4.0 mL of 6.0 M Acetic acid, 46.0 mL water and 3.3 grams of Sodium acetate trihydrate before and after 1.0 mL of 3 M NaOH is added to the solution.        Ka of HAc = 1.8 x 10 ^-5

Please Use ICE method thank you

Answer 1

pH of acetic acid + sodium acetate

pH = pKa + log(A-/HA)

pH = 4.75 + log(acetate/acid)

ICE = initially, change, Equilibrium

initially:

Vtotal = 4+46 = 52

acid = MV = 4*6/52 = 0.4615 M

acetate = mol/V = (3.3/82.0343)/(52*10^-3) = 0.77359

the change:

mmol of base added = MV = 1*3 = 3 mmol of base added

acid = -3*10^-3

base = +3*10^-3

in equilbirium

acid = 0.4615 - 3*10^-3 = 0.4585

base = (3.3/82.0343) + 3*10^-3 = 0.0432270

then

pH = 4.75 + log(acetate/acid)

pH = 4.75 + log(0.0432270/0.4585)

pH = 3.72441