Question

Enough of a monoprotic acid is dissolved in water to produce a .0167 M solution. The pH of the resulting solution is 6.03. calcualte the Ka for the acid.

OK! Im trying again to get help with this problem. I had another one like it and the way that everyone has been solving it worked for those numbers. But on this one the typical way isnt working and I dont understand how to apply what the bold red letters are saying. Please help. I have also tried 5.11x10^-11 and that was wrong as well.

Enough of a monoprotic acid is dissolved in water to produce a .0167 M solution. The pH of the resulting solution is 6.03. Calculate the Ka for the acid.

Answer 1

From the given pH = 6.03, [H₃O⁺]⁺ = 10^-6.03 = 9.332*10^-7 M

actual value of H⁺ because of the addition of [H⁺]_acid + [H⁺]_H2O = 9.332*10^-7 M

             Thus - [H⁺]_acid = (9.332*10^-7 M - 10^-7 ) = 8.332*10^-7 M

Thus -

               K_a = [H₃O⁺][A^-] / [HA] = (8.332*10^-7)(8.332*10^-7) / 0.0167 = 4.15*10^-11