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Chemistry · General chemistry II · Worked example

Find the pH of a strong acid solution

Calculate the pH, [OH⁻] and pOH of 0.0025 M hydrochloric acid at 25 °C.

A strong acid ionizes completely

Each HCl gives one H₃O⁺, so [H₃O⁺] = 0.0025 M.

Take the negative logarithm

Two significant figures in 0.0025 give two decimal places in the pH.

pH=−log(0.0025)=2.60

Find [OH⁻] from Kw

Divide Kw by [H₃O⁺].

[OH−]=1.0×10−140.0025=4.0×10−12 M
[OH−]=1.0×10−140.0025=4.0×10−12 M

Check with pOH

pOH = −log(4.0 × 10⁻¹²) = 11.40, and the two add to 14.00, as they must at 25 °C.

2.60+11.40=14.00

Result

pH = 2.60, [OH⁻] = 4.0 × 10⁻¹² M and pOH = 11.40.

Your turn

Find the pH of 0.0050 M NaOH at 25 °C.

Show the answer and explanation

pH = 11.70.

NaOH is a strong base, so [OH⁻] = 0.0050 M and pOH = −log(0.0050) = 2.30. Then pH = 14.00 − 2.30 = 11.70.

14.00−2.30=11.70

Keep exploring

In Acid, base & pH, dilute the acid to 1.0 × 10⁻⁸ M. The pH is 6.98, not 8.00: at that dilution water’s own ions matter, and adding an acid can never make water basic.

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