Question 8.E.G: The pH of 0.10 M ethylamine is 11.82. (a) Without referring ...
The pH of 0.10 M ethylamine is 11.82.
(a) Without referring to Appendix G, find K_{b} for ethylamine.
(b) Using results from part (a), calculate the pH of 0.10 M ethylammonium chloride.
(a) \underset{F – x}{CH_{3}CH_{2}NH_{2}} + H_{2}O \rightleftharpoons \underset{x}{CH_{3}CH_{2}NH_{3}^{+}} + \underset{x}{OH^{- }}
Because pH = 11.82, [OH^{−}] = K_{w}/10^{−pH} = 6.6 × 10^{−3} M = [BH^{+}].
[B] = F − x = 0.093 M.
K_{b} = \frac{[BH^{+}][OH^{- }]}{[B]} = \frac{(6.6 × 10^{−3})²}{0.093} = 4.7 × 10^{−4}(b) \underset{F – x}{CH_{3}CH_{2}NH_{3}^{+}} \rightleftharpoons \underset{x}{CH_{3}CH_{2}NH_{2}} + \underset{x}{H^{+ }}
K_{a} = \frac{K_{w}}{K_{b}} = 2.1 × 10^{−11}\frac{x²}{F − x} = K_{a} ⇒ x = 1.4_{5} × 10^{−6} M ⇒ pH = 5.84
Related Answered Questions
Question: 8.4
Verified Answer:
In Appendix G, we find that pK_{a} ...
Question: 8.3
Verified Answer:
When ammonia is dissolved in water, its reaction i...
Question: 8.2
Verified Answer:
We assume that ammonium halide salts are completel...
Question: 8.E.J
Verified Answer:
(a) pH = pK_{a} + \log \frac{[B]}{[BH^{+}]...
Question: 8.E.A
Verified Answer:
pH = −\log \mathcal{A}_{H^{+}} . Bu...
Question: 8.E.B
Verified Answer:
(a) Charge balance: [H^{+}] = [OH^{−}] + [B...
Question: 8.E.K
Verified Answer:
The reaction of phenylhydrazine with water is
[lat...
Question: 8.E.L
Verified Answer:
We use Goal Seek to vary cell B5 until cell D4 is ...
Question: 8.E.I
Verified Answer:
pH = 4.25 + log 0.75 = 4.13
Question: 8.E.H
Verified Answer:
Compound
pK_{a} (for conjugate acid...