# 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?

Jun 18, 2018

pH = 1.76.

#### Explanation:

The chemical equation is

$\text{HCOOH" + "H"_2"O" ⇌ "HCOO"^"-" + "H"_3"O"^"+}$

We can use an ICE table to solve the problem.

$\textcolor{w h i t e}{m m m m m l l} \text{HCOOH" + "H"_2"O" ⇌ "HCOO"^"-" + "H"_3"O"^"+}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m l l} 1.65 \textcolor{w h i t e}{m m m m m m m m l} 0 \textcolor{w h i t e}{m m m m} 0$
$\text{C/mol·L"^"-1": color(white)(mm)"-"xcolor(white)(mmmmmmmml)"+"xcolor(white)(mmll)"+} x$
$\text{E/mol·L"^"-1": color(white)(m)"1.65-} x \textcolor{w h i t e}{m m m m m m m m} x \textcolor{w h i t e}{m m m l l} x$

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"

$\text{pH = -log"["H"_3"O"^"+"] = "-log} 0.0172 = 1.76$