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?

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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?

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Answer 1 · 2018-06-18T00:38:58

pH = 1.76.

Explanation:

The chemical equation is

$HCOOH + H_{2} O ⇌ {HCOO}^{-} + H_{3} O^{+}$ $"HCOOH" + "H"_2"O" ⇌ "HCOO"^"-" + "H"_3"O"^"+"$

We can use an ICE table to solve the problem.

$m m m m m l l HCOOH + H_{2} O ⇌ {HCOO}^{-} + H_{3} O^{+}$ $color(white)(mmmmmll)"HCOOH" + "H"_2"O" ⇌ "HCOO"^"-" + "H"_3"O"^"+"$
${I/mol\cdotL}^{-1} : m l l 1.65 m m m m m m m m l 0 m m m m 0$ $"I/mol·L"^"-1": color(white)(mll)1.65 color(white)(mmmmmmmml)0color(white)(mmmm)0$
${C/mol\cdotL}^{-1} : m m - x m m m m m m m m l + x m m l l + x$ $"C/mol·L"^"-1": color(white)(mm)"-"xcolor(white)(mmmmmmmml)"+"xcolor(white)(mmll)"+"x$
${E/mol\cdotL}^{-1} : m 1.65- x m m m m m m m m x m m m l l x$ $"E/mol·L"^"-1": color(white)(m)"1.65-"xcolor(white)(mmmmmmmm)xcolor(white)(mmmll)x$

$K_{a} = \frac{[{HCOO}^{-}] [H_{3} O^{+}]}{[HCOOH]} = \frac{x^{2}}{1.65 - x} = 1.8 \times 10^{-4}$ $K_text(a) = (["HCOO"^"-"]["H"_3"O"^"+"])/(["HCOOH"]) = x^2/(1.65-x) = 1.8 × 10^"-4"$

Check for negligibility:

$\frac{1.65}{1.8 \times 10^{-4}} = 9200 > 400$ $1.65/(1.8 ×10^"-4") = 9200 > 400$ . ∴ x ≪ 1.65.

Then

$\frac{x^{2}}{1.65} = 1.8 \times 10^{-4}$ $x^2/1.65 = 1.8 × 10^"-4"$

$x^{2} = 1.65 \times 1.8 \times 10^{-4} = 2.97 \times 10^{-4}$ $x^2 = 1.65 × 1.8 × 10^"-4" = 2.97 × 10^"-4"$

$x = 0.0172$ $x = 0.0172$

$[H_{3} O^{+}] = 0.0172 mol/L$ $["H"_3"O"^"+"] = "0.0172 mol/L"$

$pH = -log [H_{3} O^{+}] = -log 0.0172 = 1.76$ $"pH = -log"["H"_3"O"^"+"] = "-log"0.0172 = 1.76$

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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?

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