Need help solving these two chem questions

Question

1)

In the laboratory, a general chemistry student measured the pH of a 0.482 M aqueous solution of hypochlorous acid to be 3.900.

Use the information she obtained to determine the K_a for this acid.

K_a(experiment) = 

2)In the laboratory, a general chemistry student measured the pH of a 0.482 M aqueous solution of hydrocyanic acid to be 4.875.

Use the information she obtained to determine the K_a for this acid.

K_a(experiment) = 



Answer 1

Both of these questions are solved the same way. They are the same question but using different weak acids (hypochlorous and hydrocyanic). I'll do the first one, and you can try the second one.

Hypochlorous acid, HClO:

HClO <==> H⁺ + ClO^-

Ka = [H⁺][ClO^-] / [HClO]

From the pH, we can determine the [H⁺] and [ClO^-] as follows:

Since pH = -log [H⁺], we have [H⁺] = 1x10^-3.900 = 1.26x10^-4 M = [H⁺] = [ClO^-]

We are given the [HClO] = 0.482 M, so we now have all we need to solve for Ka

Ka = (1.26x10^-4)(1.26x10^-4) / 0.482 - 1.26x10^-4) .. we can neglect 1.26x10^-4 in the denominator

Ka = 1.59x10^-8 / 0.482

Ka = 3.3x10^-9

Hydrocyanic acid, HCN

HCN <==> H⁺ + CN^-

Follow the same procedure as above