Question

Calculate the pH after 0.017 mole of NaOH is added to 1.00 L of each of the four solutions. (Assume that all solutions are at 25°C.)

a) 0.143 M acetic acid (HC₂H₃O₂, K_a = 1.8 ✕ 10⁻⁵)

b) 0.143 M sodium acetate (NaC₂H₃O₂)

c)  pure H₂O

d) 0.143 M HC₂H₃O₂ and 0.143 M NaC₂H₃O₂

Answer 1

In (a) we start with 0.143 M x 1 L = 0.143 mol of acid. You add 0.017 mol of base. Your final NaOH concentration is zero, and your final acid concentration is 0.143 - 0.017 = 0.126, and you have formed 0.017mol of the conjugate base.

pH = pKa + ([A(-) / [HA]) = 4.74 + log (0.017 / 0.126) = 4.74-0.88=3.86

In (b) we have base to start with, but its a weak base. The strong base will dominate the pH.

pOH = -log(0.017) = 1.77
pH = 14 - pOH = 14 - 1.77 = 12.23

In (c) it's the same as in (b).

In (d) only the acid will react, so you get 0.126 mol of HA and 0.163 mol of the base (some was already there, plus the stuff you form in the reaction of HA with NaOH).

pH = pKa + ([A(-) / [HA]) = 4.74 + log (0.163 / 0.126) = 4.74 + 0.11 =4.85