need help please

Question

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

Answer 1

Methanoic acid (formic acid) has a Ka = 1.8x10^-4 (Wikipedia)

Let HA = methanoic acid

HA ==> H⁺ + A^-

Ka = [H⁺][A^-] / [HA]

1.8x10^-4 = (x)(x) / 0.18 - x (assume x is small compared to 0.18 and ignore it)

1.8x10-4 = x2 / 0.18

x² = 3.24x10^-5

x = 1.8x10^-3 (since this is only 1% of 0.18, our above assumption was valid)

[H3O^+] = 1.8x10^-3 M

Answer 2

Your options of a, b, c, d, e did not post.