Question

References Use the References to access important values if needed for this question. A 28.2 mL sample of 0.304 M dimethylamine, (CH),NH, is titrated with 0.247 M hydroiodic acid. At the equivalence point, the pH is Use the Tables link in the References for any equilibrium constants that are required. Submit Answer

Answer 1

Q1. dimethylamine and hydroiodic acid

At the equivalence point, the pH is 5.82

Q2. dimethylamine and hydrobromic acid

After adding 12.6 mL of hydrobromic acid, the pH is 10.87

Explanation

Q1. moles (CH₃)₂NH = (concentration (CH₃)₂NH) * (volume (CH₃)₂NH)

moles (CH₃)₂NH = (0.304 M) * (28.2 mL)

moles (CH₃)₂NH = 8.5728 mmol

moles HI required = moles (CH₃)₂NH

moles HI required = 8.5728 mmol

volume HI required = (moles HI required) / (concentration HI)

volume HI required = (8.5728 mmol) / (0.247 M)

volume HI required = 34.7 mL

Total volume at equivalence point = (volume (CH₃)₂NH) + (volume HI)

Total volume at equivalence point = (28.2 mL) + (34.7 mL)

Total volume at equivalence point = 62.9 mL

concentration (CH₃)₂NH₂⁺ at equivalence point = (moles (CH₃)₂NH) / (Total volume at equivalence point)

concentration (CH₃)₂NH₂⁺ at equivalence point = (8.5728 mmol) / (62.9 mL)

concentration (CH₃)₂NH₂⁺ at equivalence point = 0.136 M

K_a (CH₃)₂NH₂⁺ = 1.7 x 10^-11

ICE table	(CH3)2NH2+ (aq)	$\rightleftharpoons$	(CH3)2NH (aq)	H+ (aq)
Initial conc.	0.136 M		0	0
Change	-x		+x	+x
Equilibrium conc.	0.136 M - x		+x	+x

K_a = [(CH₃)₂NH]_eq[H⁺]_eq / [(CH₃)₂NH₂⁺]_eq

1.7 x 10^-11 = [(x) * (x)] / (0.136 M - x)

solving for x, x = 1.5 x 10^-6 M

[H⁺] = x = 1.5 x 10^-6 M

pH = -log[H⁺]

pH = -log(1.5 x 10^-6 M)

pH = 5.82