Question

You start with a 150 mL solution of unknown weak monoprotic acid (HA).
a) After adding 100 mL of 0.150 M NaOH, you find half the weak acid is now deprotonated. Given that, what was the concentration of the weak acid prior to adding the NaOH?
b) After adding the 100 mL of 0.150 M NaOH to consume half the weak acid, the pH of the solution was 4.22. What was the pH of the solution before adding the NaOH?
c) What would the pH of the solution be after adding 300 mL of 0.150 M NaOH

Answer 1

(a) on adding 100ml of NaOH , half of acid is neutralized. End point reaches at 200ml .

We can apply M1V1 = M2V2

M1 × 150ml = 0.150M × 200ml

Molarity of weak monoprotic acid = 0.20M

(b)

At half equivalent point , [HA] = [A^-]

Hence pH = pKa = 4.22

HA H⁺ + A^- Ka = 10^-4.22

[H⁺] = √(c.Ka)

pH = (pKa - log c )/2 = (4.22 - log 0.2)/2 = 2.4595

pH = 2.46 (Answer)

(c)

Millimole of NaOH added = 300ml × 0.150M = 45mmol

Millimole of HA = 150ml × 0.2M = 30mmol

[OH^-] = (45-30)mmol/450ml = 0.0333M

pOH = - log 0.0333= 1.477

pH = 14 - pOH = 14 - 1.477 = 12.523

pH = 12.5 (answer)