. A 50.00 mL sample of 0.200 M HNO3 was titrated with 0.200 M NaOH. Determine the pH of the solution at the following points during the titration:

Question

. A 50.00 mL sample of 0.200 M HNO3 was titrated with 0.200 M NaOH. Determine the pH of the solution at the following points during the titration: before addition of base at the halfway point

Answer 1

Titration of a strong acid with a strong base (no buffer formed):

Before addition of base:

0.200 M HNO₃ = 0.200 M H⁺

pH = -log 0.200 = 0.699

At the halfway point:

0.05 L x 0.200 mol/L = 0.01 moles HNO₃ initially present

half titration = 0.005 moles of NaOH required since NaOH + HNO₃ ==> NaNO₃ + H₂O (1:1 mole ratio)

0.200 mol/L (x L) = 0.005 mol and x = 0.025 L = 25 ml of NaOH required for half titration

Final volume = 50.00 ml + 25.00 ml = 75 ml

[H⁺] = 0.005 mol/0.075 L = 0.0667 M

pH = -log 0.0667 = 1.18