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Chemistry · General chemistry II · Worked example

Find the pH of a weak acid

Find the pH of 0.200 M acetic acid, HC₂H₃O₂, at 25 °C. Its Ka is 1.8 × 10⁻⁵.

x20.200−x=1.8×10−5

Set up the ICE table

Acetic acid gives a proton to water: HC₂H₃O₂ + H₂O ⇌ H₃O⁺ + C₂H₃O₂⁻. Let x be the concentration that ionizes; water’s own H₃O⁺ is negligible here.

ICE table for 0.200 M acetic acid
HC₂H₃O₂ (M)H₃O⁺ (M)C₂H₃O₂⁻ (M)
Initial0.200≈ 00
Change−x+x+x
Equilibrium0.200 − xxx

Write Ka

Substitute the equilibrium row into the Ka expression.

x20.200−x=1.8×10−5

Assume x is small

Ka is tiny next to 0.200, so 0.200 − x ≈ 0.200 and x² = (1.8 × 10⁻⁵)(0.200) = 3.6 × 10⁻⁶.

x=(1.8×10−5)⁢(0.200)=1.9×10−3
x=(1.8×10−5)⁢(0.200)=1.9×10−3

Check the approximation

x is 0.95% of 0.200, well under 5%, so the approximation holds. Solving the quadratic instead gives 1.89 × 10⁻³ M, the same to two significant figures.

1.9×10−30.200×100⁢%=0.95⁢%

Find the pH

[H₃O⁺] = x.

pH=−log(1.9×10−3)=2.72

Result

[H₃O⁺] = 1.9 × 10⁻³ M and pH = 2.72; just under 1% of the acid ionizes.

Your turn

Find the pH of 0.500 M acetic acid.

Show the answer and explanation

pH = 2.52.

x ≈ √((1.8 × 10⁻⁵)(0.500)) = √(9.0 × 10⁻⁶) = 3.0 × 10⁻³ M, which is 0.60% of 0.500, so the approximation holds. pH = −log(3.0 × 10⁻³) = 2.52.

(1.8×10−5)⁢(0.500)=3.0×10−3
(1.8×10−5)⁢(0.500)=3.0×10−3

Keep exploring

In Acid, base & pH, dilute the acid tenfold, to 0.0200 M. The pH rises only to 3.23, because a larger fraction ionizes: about 3% instead of about 1%.

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