Monoprotic acid pH with a concentration of 0.022M

Question: What is the pH of a monoprotic acid with a concentration of 0.022 M?

Answer:
1. Assume this is a strong monoprotic acid so that it fully dissociates: HA → H⁺ + A^-

2. [HA] = [H⁺] = 0.022 M

3. pH = -log₁₀[H⁺] = -log₁₀[0.022] = 1.7

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