Question

1. A 23.6 mL sample of 0.391 M ethylamine, C2H5NH2, is titrated with 0.315 M nitric acid. At the equivalence point, the pH is .  

Ethylamine Kb = 4.3X10-4

Nitrous acid Ka1 = 4.5x10-4

2. A 29.1 mL sample of 0.336 M ethylamine, C₂H₅NH₂, is titrated with 0.276 M hydrochloric acid.

After adding 52.4 mL of hydrochloric acid, the pH is .

hydrochloric acid Ka1 = 3.5x10-8

Ethylamine Kb = 4.3X10-4

Answer 1

Q1. At the equivalence point, the pH is 5.70

Q2. After adding 52.4 mL of hydrochloric acid, the pH is 1.24

Explanation

Q1. concentration ethylamine = 0.391 M

volume ethylamine = 23.6 mL

moles ethylamine = (concentration ethylamine) * (volume ethylamine)

moles ethylamine = (0.391 M) * (23.6 mL)

moles ethylamine = 9.2276 mmol

moles nitric acid required to equivalence point = moles ethylamine

moles nitric acid required to equivalence point = 9.2276 mmol

volume nitric acid required = (moles nitric acid required to equivalence point) / (concentration nitric acid)

volume nitric acid required = (9.2276 mmol) / (0.315 M)

volume nitric acid required = 29.3 mL

Total volume at equivalence point = (volume ethylamine) + (volume nitric acid)

Total volume at equivalence point = (23.6 mL) + (29.3 mL)

Total volume at equivalence point = 52.9 mL

concentration of C₂H₅NH₃⁺ at equivalence point = (moles ethylamine) / (Total volume at equivalence point)

concentration of C₂H₅NH₃⁺ at equivalence point = (9.2276 mmol) / (52.9 mL)

concentration of C₂H₅NH₃⁺ at equivalence point = 0.174 M

K_b ethylamine = 4.3 x 10^-4

K_a C₂H₅NH₃⁺ = (K_w) / (K_b)

K_a C₂H₅NH₃⁺ = (1.0 x 10^-14) / (4.3 x 10^-4)

K_a C₂H₅NH₃⁺ = 2.3 x 10^-11

ICE table	C2H5NH3+ (aq)	$\rightleftharpoons$	C2H5NH2 (aq)	H+ (aq)
Initial conc.	0.174 M		0	0
Change	-x		+x	+x
Equilibrium conc.	0.174 M - x		+x	+x

K_a = [C₂H₅NH₂]_eq[H⁺]_eq / [C₂H₅NH₃⁺]_eq

2.3 x 10^-11 = [(x) * (x)] / (0.174 M - x)

Solving for x, x = 2.0 x 10^-6 M

[H⁺] = x = 2.0 x 10^-6 M

pH = -log[H⁺]

pH = -log(2.0 x 10^-6 M)

pH = 5.70