What is the pH of a solution prepared by dissolving 0.080 mol of a strong monoprotic acid HA in enough water to make 400.0 mL of solution? Assume complete dissociation.
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Correct answer
A. 0.70
Principle or equation
For a strong monoprotic acid, [H+] = initial acid concentration. pH = -log[H+].
Why this answer is correct
Molarity = 0.080 mol / 0.400 L = 0.20 M. Since HA is strong, [H+] = 0.20 M. pH = -log(0.20) = 0.70.
Example
If 0.010 mol of HCl is in 0.100 L, [H+] = 0.10 M, pH = 1.00.
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