Question

Find the pH during the titration of 20.00 mL of 0.1590 M nitrous acid, HNO2 (Ka = 7.1 ✕ 10-4), with 0.1590 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

a) We have only nitrous acid, we make our ICE table and establish dissociation equation:

HNO₂ <-> NO₂^- + H₃O⁺

	HNO2	<->	NO2-	+	H3O+
I	0.159		0		0
C	-x		+x		+x
E	0.159 - x		x		x

We use constant of acid definition:

Ka = [NO₂^-] [H₃O⁺] / [HNO₂]

7.1 x 10^-4 = x² / (0.159 - x)

Isolation for x:

x = [H⁺] = 0.0103 M

pH = -log(H⁺) = 1.9872

b) When adding 10 mL, we have:

0.01 L * 0.159 mol/L = 0.00159 moles of NaOH

0.02 L * 0.159 mol/L = 0.00318 moles of HNO₂

Half the moles will be consumed, so we'll have 0.00159 moles of HNO2 remaining and:

NaOH + HNO₂ -> H₂O + NaNO₂

We have 0.00159 moles of basic salt, so we determine pH with the following expression:

pH = pKa + log [Base/Acid]

pH = -log (7.1 x 10^-4) + log [1]

pH = 3.1487

c) Adding 15 mL, we'll be adding:

0.015 L * 0.159 mol/L = 0.00253 moles of NaOH

Leaving:

0.00318 - 0.00253 moles of HNO₂ = 0.000795 of HNO₂

And having now 0.00253 moles of base

We obtain pH:

pH = pKa + log [Base/Acid]

pH = -log (7.1 x 10^-4) + log [0.00253/0.000795]

pH = 3.6515

d) Adding 19 mL, we'll be adding:

0.019 L * 0.159 mol/L = 0.003021 moles of NaOH

Leaving:

0.00318 - 0.003021 moles of HNO₂ = 0.000159 of HNO₂

And having now 0.003021 moles of base

We obtain pH:

pH = pKa + log [Base/Acid]

pH = -log (7.1 x 10^-4) + log [0.003021/0.000159]

pH = 4.4275

e) Adding 19.95 mL, we'll be adding:

0.01995 L * 0.159 mol/L = 0.003172 moles of NaOH

Leaving:

0.00318 - 0.003172 moles of HNO₂ = 0.00000795 of HNO₂

And having now 0.003172 moles of base

We obtain pH:

pH = pKa + log [Base/Acid]

pH = -log (7.1 x 10^-4) + log [0.003172/0.00000795]

pH = 5.7497

f) Adding 20 mL will consume all acid and we'll be having only basic salt. So we obtain Kb:

Kb = Kw / Ka = 1 x 10^-14 / 7.1 x 10^-4 = 1.4 x 10^-11

We do then the following:

[OH^-] = sqrt * (Kb * [Salt]) = sqrt((1.4 x 10^-11) * (0.00318 moles/0.04L))

[OH-] = 1.055 x 10^-6 M

pOH = -log (1.055 x 10^-6) = 5.9767

pH = 14 - 5.9767 = 8.023

g) We'll now be having strong base, so we only get concentration of strong base:

Moles: 0.00005 L * 0.159 M = 0.00000795 moles

Volume = 40.05 / 1000 = 0.0405 L

[OH^-] = 0.00000795 / 0.0405 = 1.963 x 10^-4 M

pOH = -log(1.963 x 10^-4) = 3.7

pH = 14 - 3.7 = 10.29

h) Same case:

Moles: 0.005 L * 0.159 M = 0.000795 moles

Volume = 45 / 1000 = 0.045 L

[OH^-] = 0.000795 / 0.045 = 0.01767 M

pOH = -log(0.01767) = 1.7528

pH = 14 - 1.7528 = 12.2471