Closing Calculations

In your calculations, the preliminary focus of the weak acid is:

M

The Ka is:

1.810^-4

Within the equilibrium relations of weak acids, the x = [H ] = Ka x Focus = 1.810^-4 x = 6.410^-3

pH = -log[H ] = -log (6.410^-3) = 2.19

Then:

[H ] = 10-pH = 10-(2.19) = 6.510^-3

HCOOH H2O <> HCOO- H3O

Ka(HCOOH) = 1.8E-4

Make an ICE desk (I will not do it right here to avoid wasting time).

Ka = [H3O ][HCOO-] / [HCOOH]

Ka = (x)(x) / (0.230-x) = 1.8E-4

1.8E-4 = x / (0.230-x)

If we assume 0.230>>x, then

1.8E-4 = x / 0.230

x = 0.0064342832M = [H3O ]

[Answer: see above]

This post is last updated on hrtanswers.com at Date : 1st of September – 2022