A solution contains 0.50 (K_a = 2.0 × 10^-8) and 0.22 M NaA. Calculate the pH after 0.05mol of NaOH is added to 1.00 L of this solution. A) 7.48 B) 7.70 C) 7.34 D) 7.92 E) 7.19

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Quizplus · Accepted Answer

The addition of NaOH to the solution will consume the HA (acid form) and generate A- (conjugate base), shifting the equilibrium according to the reaction HA + OH- -> A- + H2O. Since NaOH is a strong base, it will completely dissociate and react with HA. The amount of HA decreases by 0.05 mol, and the amount of A- increases by 0.05 mol. The new concentrations in the 1.00 L solution are [HA] = 0.50 M - 0.05 M = 0.45 M and [A-] = 0.22 M + 0.05 M = 0.27 M. Using the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), where pKa = -log(Ka) = -log(2.0 × 10^-8) = 7.70. Substituting the new concentrations, pH = 7.70 + log(0.27/0.45) = 7.70 - 0.22 = 7.48.

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