In the laboratory, a general chemistry student measured the pH of a 0.553 M aque

Question

In the laboratory, a general chemistry student measured the pH of a 0.553 M aqueous solution of triethanolamine, CoH1sO3N to be 10.766. Use the information she obtained to determine the K, for this base. Ks(experiment) - In the laboratory, a general chemistry student measured the pH of a 0.553 M aqueous solution of trimethylamine, (CH3)3N to be 11.754. Use the information she obtained to determine the K, for this base. Kb(experiment) The pH of an aqueous solution of 0.553 M triethanolamine (a weak base with the formula C&HisO3N;) is The hydronium ion concentration of an aqueous solution of 0.553 M trimethylamine (a weak base with the formula (CH3)3N) is

Explanation / Answer

1) [base] = 0.553M

pH of solution =10.766

We know that pH + pOH = 14

thus pOh of this solution = 3.234

and pOH of a weak base is

pOH = 1/2[pKb- log C]

3.234 = 1/2 [pkb -log 0.553]

Kb = 6.15x10-7

2) Here

[base] = 0.553M

pH of solution =11.754

We know that pH + pOH = 14

thus pOh of this solution = 2.246

and pOH of a weak base is

pOH = 1/2[pKb- log C]

2.246= 1/2 [pkb -log 0.553]

Kb = 5.82x10-5

3)The pKb of triethanolamine is 6.26 (from wikipedia)

[base] = 0.553 M

pOH = 1/2[pKb -logc]

= 1/2 [6.26 -log 0.553]

= 3.259

pH = 14-3.259

= 10.74

4) We have the [Oh-] in trimethylamine as

[OH-] = square root of kb xc

= square root of (6.46x10-5x 0.553)

= 5.97x10-2 M

Thus hydronium ion [H3O+] = 1.0x10-14 /5.97x10-2 M

=1.675x10-13 M

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