Question

Find the pH during the titration of 20.00 mL of 0.2350 M butanoic acid, CH₃CH₂CH₂COOH (K_a = 1.54 ✕ 10^-5), with 0.2350 M NaOH solution after the following additions of titrant.

(a)    0 mL


(b)    10.00 mL


(c)    15.00 mL


(d)    19.00 mL


(e)    19.95 mL


(f)    20.00 mL


(g)    20.05 mL


(h)    25.00 mL

Answer 1

At 0 ml NaOH:
0.2350 M HA
Ka = 1.54x10^-5 = [H+][A-]/[HA]
let X = [H+]
1.54x10^-5 = [H+][A-]/[HA] = X^2 / (0.2350 - X)
solving cuadratic equation:
X = 0.001895 M
pH = 2.72

At 10.0, 15.00, 19.00, and 19.95 mL NaOH added you have a buffer solution composed of HA and A^-1. Use the Henderson-Hasselbalch equation-
pH = pKa + log(A-/HA)

at 10.00 mL NaOH added HA=A- so pH = pKa
pH = -log(1.54x10^-5) = 4.81

at 15.00 mL NaOH added, 3/4 of HA has been converted to A-
pH = pKa + log(0.75/0.25) = 5.29

at 19.00 mL NaOH added
started with 4.7 millimoles of HA (20.00 mL x 0.2350 M = 4.7 mmol)
added 4.465 millimol NaOH (19.00 mL x 0.2350 M = 4.465)
now have 4.465 millimol A- and 0.2235 millimol HA
pH = pKa + log(4.465/0.235) = 6.09

You do the next for 19.95 mL NaOH added

at 20.00 mL NaOH added you have a 40.00 mL of a solution containing 5.34 millimoles of A-
A- + H2O <--> HA + OH-
Kb = [HA][OH-]/[A-]
[A-] = 5.34 mmol / 40.00 mL = 0.1335 M
Kb = Kw/Ka = 1x10^-14/1.54x10^-5 = 6.49x10^-10
let X = [OH-]
Kb = 6.49x10^-10 = X^2 / (0.1335 - X)
X = 9.31x10^-6
pOH = 5.03
pH = 14.00 - 5.03 = 8.97

at 20.05 mL and 25.00 mL of NaOH the excess NaOH is the main source of OH- and not the equilibrium A- + H2O <--> HA + OH-

At 20.05 mL NaOH added, you have added 0.05 mL excess NaOH
0.05 x 0.2670 M = 0.01335 millimoles
0.01335 millimoles / 40.05 mL = 0.000333 M
pOH = 3.48
pH = 14.00 - 3.48 = 10.52

You can calculate the pH after 25.00 mL of NaOH are added