What is the pH of a solution prepared by dissolving 0.0250 mol of a strong monoprotic acid HA in enough water to make 500.0 mL of solution? Assume complete dissociation.
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Correct answer
A. 1.30
Principle or equation
For a strong monoprotic acid, [H+] = initial concentration of acid. pH = -log[H+].
Why this answer is correct
Moles of HA = 0.0250 mol, volume = 0.500 L, so [HA] = 0.0250 / 0.500 = 0.0500 M. Since it is a strong acid, [H+] = 0.0500 M. pH = -log(0.0500) = 1.301, approximately 1.30.
Example
If 0.010 mol of HCl is dissolved in 1.00 L, [H+] = 0.010 M, pH = 2.00.
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