Question

A 25.0-mL sample of 0.10 M weak base is titrated with 0.15 M strong acid.

What is the pH of the solution after 9.00 mL of acid have been added to the weak base?

Weak base K_b = 6.5 × 10^–4

Answer 1

Answes. Number of mole of weak base (B2) = 25.0mm2 x 0.10M = 0.025 L x 0.10 mel = 0.0025 mol Number of mole of strong Acid CHA) - g.oome x 0.ISMO - 0.oogtx Oois mol = 0.00135 mol t Pkb of weak base, pkb = -log (K6) Pkb = – log (6.5X104) Pko = {log 6.5t log lot) pk = - CO-8129-4) pkb = -(-3.187) pkb = 30187 Chemical Equation, B THA - BHT +A- weak base stong Acid consugate conjugate Base e.g. weak Basc, B. B +H20 - But toh 0.0025 – 0 -0.00135 to-00135 + 0.00135 o 0.oolis 0.00135 0.00135 Aced

Because g.oome of and have been added, and we started 25.0m2 Base, There are Total volume of concelente es 34.ome of Solution, Therefore, Molanity of Base (6) = 0.0025 mol = 0.0025 mol 340ml Solution 0.0342 = 0.0735M 1 molarity of (BH+) = 0.00135 mol = 0·034 L 0.0397M Now, we can used approximation, Henderson- Hasselbalch por= pkb + log [Bht) [ B] DOM = POH = 3.187 + log [0.0397] [0.0735] 3.187 +/log [0.0397) -log[0.0735]) pot = 3.187 +[-1.401 +1.133] pot = 3.187 – 0.268 en c an pH = 14 - POH! ph= 14 – 2.919. TpH = 11.08