The [H3O+] in a solution of 0.18 methanoic acid?

Question

The [H₃O⁺] in a solution of 0.18 methanoic acid is a.bcdef? mol/L

Answer 1

You need to know the Ka for methanoic acid (formic acid). This can be found in a text book, or on the internet. The value I found is 1.8x10^-4.

In the following analysis, H⁺ = H₃O⁺ and H²O is omitted for clarity.

HCOOH ==> H⁺ + HCOO^-

Ka = [H⁺][HCOO^-]/[HCOOH]

1.8x10^-4 = (x)(x)/0.18 - x (assuming x is small relative to 0.18, we can omit it from the denominator)

1.8x10^-4 = x²/0.18

x² = 3.24x10^-5

x = 1.8x10^-3 M = [H⁺] (note that this calculated value is indeed small relative to 0.18. It is 1/1000 the amount)