Calculate the pH of the solution resulting from the addition of 10.0 mL of 0.10 M NaOH to 50.0 mL of 0.10 M HCN (K_a = 4.9 * 10^-10) solution.

Question

Quizplus · Accepted Answer

The pH of the solution can be calculated by first determining the amount of HCN neutralized by NaOH, then calculating the concentration of the resulting CN- ion, and finally using the Kb of CN- to find the pH. HCN is a weak acid and NaOH is a strong base. The reaction between HCN and NaOH will produce water and CN- (the conjugate base of HCN). The initial moles of HCN are 0.0050 mol (50.0 mL * 0.10 M) and the moles of NaOH are 0.0010 mol (10.0 mL * 0.10 M). Since NaOH is limiting, all of it will react, leaving 0.0040 mol of HCN and producing 0.0010 mol of CN-. The total volume after reaction is 60.0 mL. The concentration of CN- is 0.0010 mol / 0.060 L = 0.0167 M. Using the Kb for CN- (Kw/Ka for HCN, where Kw = 1.0 x 10^-14 and Ka for HCN = 4.9 x 10^-10), Kb = 1.0 x 10^-14 / 4.9 x 10^-10 = 2.04 x 10^-5. Applying the formula for Kb, [OH-] can be calculated and then pOH, and finally pH. The calculation shows that the pH is approximately 8.71, indicating a basic solution as expected from the reaction of a weak acid with a strong base.

Calculate the PH of the Solution Resulting from the Addition