A student has 0.01 mol of a weak monoprotic organic acid. He prepares 100 mL of its aqueous solution. He then titrates 50 mL of this solution with NaOH to the equivalence point, and mixes the resulting solution back with the remaining 50 mL of untreated acid solution. The pH of the final m

Question

A student has 0.01 mol of a weak monoprotic organic acid. He prepares 100 mL of its aqueous solution. He then titrates 50 mL of this solution with NaOH to the equivalence point, and mixes the resulting solution back with the remaining 50 mL of untreated acid solution. The pH of the final mixture is found to be 4.80. What is the pKa of the acid (to the nearest integer)?

Accepted Answer

5. The 100 mL solution contains 0.01 mol acid (0.1 M). The 50 mL portion contains 0.005 mol acid; after titrating to equivalence with NaOH, all acid is converted to conjugate base A⁻ (0.005 mol). When mixed with the remaining 50 mL containing 0.005 mol acid HA, the final solution has equal moles of HA and A⁻. By Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]) = pKa + log(1) = pKa. So pKa = 4.80, which rounds to 5.

Solution

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