Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0122 M solution. The pH of the resulting solution is 2.64. Calculate the Ka for the acid.

Answer #1

concentration = 0.0122 M

pH of solution = 2.64

pH = -log [H+]

[H+] = 10^-2.64

[H+] = 2.29 x 10^-3 M

HA --------------------> H+ + A-

0.0122 - 2.29 x 10^-3 2.29 x 10^-3 2.29 x 10^-3

Ka = [H+][A-] / [HA]

= (2.29 x 10^-3)^2 / 0.0122 - 2.29 x 10^-3

= 5.29 x 10^-4

**Ka of acid = 5.29 x 10^-4**

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Hi can someone please help me do this question with explnation
and step by step.
Thank you

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