Titration of HNO2 with NaOH...need pH

Question

Consider a titration of 10.0 mL of 0.30 M HNO₂ with 0.10 M NaOH.

K_a HNO₂ = 7.1 x 10^-4

a. pH of solution before titration begins?

b. pH when 12.00 mL NaOH has been added?

c. pH when 15.00 mL NaOH is added

Answer 1

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Now, for the current problem, we are dealing with the titration of a weak acid (HNO₂) with a strong base (NaOH). We are asked to find the pH at different points during the titration.

(a) pH before the titration begins. This will be the pH of the 0.30 M HNO₂. We calculate that as follows:

HNO₂ <==> H⁺ + NO₂^-

Ka = 7.1 x 10^-4 = [H⁺][NO₂^-] / [HNO₂]

7.1 x 10^-4 = (x)(x) / 0.30 - x and we'll assume x is small relative to 0.30 M and ignore it

7.1 x 10^-4 = x² / 0.30

x² = 2.13x10^-4

x = [H⁺] = 0.0146 M (note this is ~ 4.8% of 0.30 M so is less than 5% and we can ignore it as assumed)

pH = -log [H⁺] = -log 0.0146

pH = 1.84

(b) pH after 12.0 ml of 0.10 M NaOH is added

initial moles HNO₂ = 10.0 mls x 1 L /1000 mls x 0.30 mol/L = 0.003 moles

moles of NaOH added = 12.0 mls x 1L/1000 mls x 0.10 mol/L = 0.0012 moles

final volume = 10.0 mls + 12.0 mls = 22.0 mls = 0.022 L

HNO₂ + NaOH ===> NaNO₂ + H₂O

0.003........0.0012.............0.................Initial

-0.0012...-0.0012..........+0.0012..........Change

0.0018.........0.................0.0012..........Equilibrium

At this point in the titration, we have both the weak acid and the conjugate base present, so we have formed a buffer. We will use the Henderson Hasselbalch equation. The pKa = -log Ka = -log 7.1 x 10^-4 = 3.15

pH = pKa + log [conj.base]/[acid] = 3.15 + log (0.0012/0.0018)

pH = 3.15 - 0.18

pH = 2.97

(c) pH after 15.0 ml of 0.10 M NaOH is added.

moles NaOH added = 15.0 mls x 1L/1000 ml x 0.10 mol/L = 0.0015 moles

HNO₂ + NaOH ===> NaNO₂ + H₂O

0.003.....0.0015............0..................Initial

-0.0015...-0.0015........+0.0015........Change

0.0015........0...............0.0015.........Equilibrium

We still have both the weak acid and the conjugate base present, so we still have a buffer solution. NOTE that the [HNO₂] = [NO₂^-] = 0.0015 M, so the pH = pKa

pH = 3.15