Question

A buffer solution contains 0.347 M KH₂PO₄ and 0.269 M Na₂HPO₄. Determine the pH change when 0.079 mol NaOH is added to 1.00 L of the buffer.

pH after addition − pH before addition = pH change =

Answer 1

Phosphoric acid hans pka values a n d tour offront pka (1) = 2.14 H₂PO4/H₂PO4 pka (2) = 7.2 H₂poa / HP09 pko (3) = 12.37 APO / POM / We will use pka(2), because the buffer solution given consist of H₂PO4 & 400 2- sobota Now, we know, the Henderson's equation for buffer solution: at pH = pka Isalt] Clive Tacid] + кути? log ory Here, pH = pka (2) [HPO 2 -] Brou & log. LOH = 17.2. t log 0.347 L and of IL buffer contains 0.347 (M) H₂por & 0.269 (M) HPO42- 20.269 so, initial, pH = 17.2. & log & = 7.0894 omto Now, after addition of 0.079 mol NaOH, PH : 1L 0.347 (M) H₂PO4 = 0.347 mol H₂PO4 ZIL 0.269 (M) HPO4 = 0.269 mol HPO 4² 0.079 mol NaOH will conssime 0.079 mol H₂POg & enfortit will produce 0.079 mol HPO 4 ² 4 so, tinal H₂PO4 = 0.347 – 0.07% = 0.268 To final HPO 42 = 0.269 + 0.079 = 0.348 en 0:348 = 7.313 T pH = 7.2 + log o 1268 15.11 pH change = 7.313 - 7.0894 and = 0.2236.