I need help solving for the Ka for this acid?

Question

You prepared an aqueous solution of a monoprotic acid (formula mass = 153.482g/mol) by dissolving 26.816g of the acid in sufficient water to make 500.0mL of solution. The pH of the resulting solution is 4.274. What is Ka for this acid?

Answer 1

Let the monoprotic acid be represented by HA

HA ==> H⁺ + A^-

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

Initial [HA] = 26.816 g x 1 mol / 153.482 g = 0.17472 mol / 0.5000 L = 0.34944 M

From the pH of 4.274 we can calculate the [H⁺] and [A^-]

[H+] = 1x10^-pH = 1x10^-2.274

[H+] = 5.321x10^-3 M

Substituting values into the Ka expression and solving for Ka, we have

Ka = (5.321x10^-3 )(5.321x10^-3 ) / 0.34944

Ka = 8.102x10^-5