A 0.94 M aqueous solution of a weak base has a pH of 10.28 at 298...

Question

Question

A 0.94 M aqueous solution of a weak base has a pH of 10.28 at 298 K. Calculate pK_b for this weak base. Enter your answer to 2 decimal places.

Answer 1

Answer #1

the ionization equation of weak base B is as following

B + H2O ⇌ HB+ + OH- and Kb is dissociation constant for the same.

now we know that

pH = -log[H+]

hence [H+] = 10-pH = 10-10.28 = 5.2481 x 10-11 M

[OH-] = Kw / [H+] = 1.00 x 10-14 / 5.2481 x 10-11

[OH-] = 1.9055 x 10-4 M

due to the 1:1 molar ratio that [HB+] = [OH-]

and we have 0.94 M for [B].

Kb = [HB+][OH-]/[B]

= [(1.9055 x 10-4) (1.9055 x 10-4)] / 0.94

Kb = 3.86 x 10-8

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Answer 2

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