Question

Find the pH during the titration of 20.00 mL of 0.1000 M nitrous acid, HNO2 (Ka = 7.1 10-4), with 0.1000 M NaOH solution after the following additions of titrant. Find at volumes of 0 mL, 10.00 mL, 15.00 mL, 20.00 mL, and 25.00 mL.

Answer 1

pH during the titration of 20.00 mL of 0.1000 M nitrous acid:

HNO₂(aq)       ↔ H⁺(aq) + NO₂^-(aq)
0.10-x M ........    x M .......x M

K_a = [H⁺] [A^-] / [HA]

7.1x10^-4 = (x)(x) / (0.10 - x)

Since the amount of x is very small compared to 0.10 M, it is neglected.
7.1x10^-4 = (x)(x) / (0.10)
x = [H⁺] = 0.842 x 10^-2M

pH = -log (0.842 x 10^-2) = 2.07

pH during the titration of 20.00 mL of 0.1000 M nitrous acid with 10.00 mL of 0.1000 M NaOH:

moles of nitrous acid = 0.020 L x (0.1 mol/L) = 0.002 moles

moles of NaOH = 0.010 L x (0.1 mol/L) = 0.001 moles

therefore [H⁺] = (0.002-0.001) moles/0.030L = 0.0333 M

pH = -log (0.0333) = 1.47

pH during the titration of 20.00 mL of 0.1000 M nitrous acid with 15.00 mL of 0.1000 M NaOH:

moles of nitrous acid = 0.020 L x (0.1 mol/L) = 0.002 moles

moles of NaOH = 0.015 L x (0.1 mol/L) = 0.0015 moles

therefore [H⁺] = (0.002-0.0015) moles/0.035L = 0.0142 M

pH = -log (0.0142) = 1.847

pH during the titration of 20.00 mL of 0.1000 M nitrous acid with 20.00 mL of 0.1000 M NaOH:

moles of nitrous acid = 0.020 L x (0.1 mol/L) = 0.002 moles

moles of NaOH = 0.020 L x (0.1 mol/L) = 0.002 moles

moles of acid = moles of base, so the reaction neutralized and the pH = 7

pH during the titration of 20.00 mL of 0.1000 M nitrous acid with 25.00 mL of 0.1000 M NaOH:

moles of nitrous acid = 0.020 L x (0.1 mol/L) = 0.002 moles

moles of NaOH = 0.025 L x (0.1 mol/L) = 0.0025 moles

therefore [OH^-] = (0.0025-0.002) moles/0.035L = 0.011 M

pOH = -log (0.011) = 1.96

pH = 14 – 1.96 = 12.04