Determine the [H3O+] of a 0.230 M solution of formic acid?

Closing Calculations
In your calculations, the preliminary focus of the weak acid is:
M
The Ka is:
1.8×10^-4
Within the equilibrium relations of weak acids, the x = [H+] = √Ka x Focus = √1.8×10^-4 x = 6.4×10^-3
pH = -log[H+] = -log (6.4×10^-3) = 2.19
Then:
[H+] = 10-pH = 10-(2.19) = 6.5×10^-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 = x² / (0.230-x)
If we assume 0.230>>x, then
1.8E-4 = x² / 0.230
x = 0.0064342832M = [H3O+]
[Answer: see above]

Also Read :   If 27.0 g of MgSO4.7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?

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