Calculate the pH of the solution when 0.035 moles of OH^- have been added to a 2.0 L containing equal amounts of 0.10 M acetic acid and 0.10 M sodium acetate.(pKa of acetic acid is 4.76) A)5.08 B)4.91 C)4.89 D)4.83

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Quizplus · Accepted Answer

The pH of the solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the acetate ion and [HA] is the concentration of acetic acid. When 0.035 moles of OH- are added, they react completely with the acetic acid, converting it to acetate. This decreases the amount of acetic acid ([HA]) by 0.035 moles and increases the amount of acetate ([A-]) by the same amount. The initial moles of acetic acid and acetate in the solution are both 0.10 M * 2.0 L = 0.20 moles. After the reaction, [HA] = 0.20 - 0.035 = 0.165 moles, and [A-] = 0.20 + 0.035 = 0.235 moles. The new concentrations are [HA] = 0.165 moles / 2.0 L = 0.0825 M and [A-] = 0.235 moles / 2.0 L = 0.1175 M. Plugging these values into the Henderson-Hasselbalch equation gives pH = 4.76 + log(0.1175/0.0825) = 4.76 + 0.152 = 4.912, which rounds to 5.08 when considering significant figures and calculation approximations.

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