Question

Find the pH during the titration of 20.00 mL of 0.1820 M nitrous acid, HNO2 (Ka = 7.1 ✕ 10-4), with 0.1820 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

pH during the titration of 20.00 mL of 0.182 M nitrous acid:

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

K_a = [H⁺] [A^-] / [HA]

7.1x10^-4 = (x)(x) / (0.182 - x)

Since the amount of x is very small compared to 0.182 M, it is neglected.
7.1x10^-4 = (x)(x) / (0.182)
x = [H⁺] = 1.136 x 10^-2M

pH = -log (1.136 x 10^-2) = 1.944

pH during the titration of 20.00 mL of 0.182 M nitrous acid with 10.00 mL of 0.182 M NaOH:

moles of nitrous acid = 0.020 L x (0.182 mol/L) = 0.00364 moles

moles of NaOH = 0.010 L x (0.182 mol/L) = 0.00182 moles

therefore [H⁺] = (0.00364-0.00182) moles/0.030L = 0.0606 M

pH = -log (0.0606) = 1.217

pH during the titration of 20.00 mL of 0.182 M nitrous acid with 15.00 mL of 0.182 M NaOH:

moles of nitrous acid = 0.020 L x (0.182 mol/L) = 0.00364 moles

moles of NaOH = 0.015 L x (0.182 mol/L) = 0.00273 moles

therefore [H⁺] = (0.00364-0.00273) moles/0.035L = 0.026 M

pH = -log (0.026) = 1.58

pH during the titration of 20.00 mL of 0.182 M nitrous acid with 19.00 mL of 0.182 M NaOH:

moles of nitrous acid = 0.020 L x (0.182 mol/L) = 0.00364 moles

moles of NaOH = 0.019 L x (0.182 mol/L) = 0.00346 moles

therefore [H⁺] = (0.00364-0.00346) moles/0.039L = 0.00461 M

pH = -log (0.00461) = 2.33

pH during the titration of 20.00 mL of 0.182 M nitrous acid with 19.95 mL of 0.182 M NaOH:

moles of nitrous acid = 0.020 L x (0.182 mol/L) = 0.00364 moles

moles of NaOH = 0.01995 L x (0.182 mol/L) = 0.00363 moles

therefore [H⁺] = (0.00364-0.00363) moles/0.03995L = 2.5 x10^-4 M

pH = -log (2.5 x10^-4) = 3.60

pH during the titration of 20.00 mL of 0.182 M nitrous acid with 20.00 mL of 0.182 M NaOH:

moles of nitrous acid = 0.020 L x (0.182 mol/L) = 0.00364 moles

moles of NaOH = 0.020 L x (0.182 mol/L) = 0.00364 moles

k_b = k_w/k_a = 10^-14/7.1 x 10^-4 = 0.14 x 10^-10

k_b = x²/ 0.00364

x² = (0.14 x 10^-10) (0.00364)

x = [OH^-] = 0.0225 x 10^-5

pOH = -log (0.0225 x 10^-5) = 3.53

pH = 14 – 3.53 = 10.47

pH during the titration of 20.00 mL of 0.182 M nitrous acid with 20.05 mL of 0.182 M NaOH:

moles of nitrous acid = 0.020 L x (0.182 mol/L) = 0.00364 moles

moles of NaOH = 0.02005 L x (0.182 mol/L) = 0.0036491 moles

therefore [OH^-] = (0.0036491-0.00364) moles/0.03995L = 2.27 x10^-4 M

pOH = -log (2.5 x10^-4) = 3.60

pH = 14 – 3.6 = 10.4

pH during the titration of 20.00 mL of 0.182 M nitrous acid with 25.00 mL of 0.182 M NaOH:

moles of nitrous acid = 0.020 L x (0.182 mol/L) = 0.00364 moles

moles of NaOH = 0.025 L x (0.182 mol/L) = 0.00455 moles

therefore [OH^-] = (0.00455-0.00364) moles/0.045L = 0.0202 M

pOH = -log (0.0202) = 1.69

pH = 14 – 1.69= 12.31