Question

Formic (or methanoic) acid is a weak acid secreted by ants as a defence mechanism. The acid has a Ka value of 1.8 x 10-4. What is the pH of a 1.65M solution of formic acid?

Answer 1

pH = 1.76.

Explanation:

The chemical equation is

`"HCOOH" + "H"_2"O" ⇌ "HCOO"^"-" + "H"_3"O"^"+"`

We can use an ICE table to solve the problem.

`color(white)(mmmmmll)"HCOOH" + "H"_2"O" ⇌ "HCOO"^"-" + "H"_3"O"^"+"`
`"I/mol·L"^"-1": color(white)(mll)1.65 color(white)(mmmmmmmml)0color(white)(mmmm)0`
`"C/mol·L"^"-1": color(white)(mm)"-"xcolor(white)(mmmmmmmml)"+"xcolor(white)(mmll)"+"x`
`"E/mol·L"^"-1": color(white)(m)"1.65-"xcolor(white)(mmmmmmmm)xcolor(white)(mmmll)x`

`K_text(a) = (["HCOO"^"-"]["H"_3"O"^"+"])/(["HCOOH"]) = x^2/(1.65-x) = 1.8 × 10^"-4"`

Check for negligibility:

`1.65/(1.8 ×10^"-4") = 9200 > 400`. ∴ x ≪ 1.65.

Then

`x^2/1.65 = 1.8 × 10^"-4"`

`x^2 = 1.65 × 1.8 × 10^"-4" = 2.97 × 10^"-4"`

`x = 0.0172`

`["H"_3"O"^"+"] = "0.0172 mol/L"`

`"pH = -log"["H"_3"O"^"+"] = "-log"0.0172 = 1.76`