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A weak acid has Ka equals 0.0003887 What is the pH of a solution of this acid when HA equals 10.92 mM and A - equals 8.07 mM?
Easyas123abc
pH = -log10(Ka) - log10([HA]/[A-]) = 3.41 -log10(1.35) = 3.41 - 0.1313 = 3.28
A line of reasoning showing more steps is: By definition, Ka = [H+][A-]/[HA], where the brackets indicate molar concentrations. Dividing both sides of this equation by [A-]/[HA] and substituting the numerical values given results in [H+] = (10.92 X 10-3)(3.887 X 10-4)/(8.07 X 10-3) = 5.260 X 10-4. The log of 5.260 =0.7210, so the pH of this solution is - (-4 - 0.7210) = 3.28, to the justified number of significant digits (limited by 8.07).
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