Part A Determine the [H3O+] of a 0

Part A:

             HCOOH < -------------->      COO-(aq) + H3O+(aq)

I              0.250                                0                0

C               -x                                   +x              +x

E            0.250-x                               +x             +x

Ka = [H3O+][COO-] / [HCOOH] = 1.8 x 10^-4

1.8 x 10^-4 = x^2 / (0.250-x)

x^2 + (1.8 x 10^-4)x - (4.5 x 10^-5) = 0

Solving the quadratic equation,

x = 0.0066

Thus, [H3O+] = x = 0.0066 M

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Part B:

pH = -log[H3O+]

pH = -log(0.0066)

pH = 2.18

Given, 0.250 M of HCOOH (formic acid)

[H3O+]