Question

How much volume of a 0.102 M NaOH solution must be added to 25.00 ml of a 0.075 M solution of benzene-1,2,3-tricarboxylic acid to obtain a pH equal to 4.80? pKa values: pK_a1 = 2.86, pk_a2 = 4.30 and pk_a3 = 6.28.

Answer 1

مکہ الحمم we obtain a pH equal to 4.80 first need to reach the first equivalence paint and then add more NaOH. so that pH becomes 4.80. Now, volume of NaOH required to reach first equivalence point is: QlO2M X Una oH = 25 ml x 0.075M » VNGOH = 254 0.075 ml 2 واوو 18.382 ml additional Now, det at mnoles of NaOH be added. HbCg H306 + NAOH → HCgH3062 + H2O Initial 25X0.075M ammoles ammoles moles 1.875 mmoles findl 10875 -X ommales ammoles mmolles Sog pH Pkaz & lage [uca H₂O6²" Cho Cg H₂06- 4.80 = 4.30 + lagro Ny mmoles 1:475-3/4 100.50 3.1623 a 1875-0%

so, 2= 1.875+ 3.1623 (301623+1 2 1.437 m moles Sor additional volume of No OH required = 10437 m moles 0.102 M 14.088 ml So, total volume of Na OH required = 18.382 + 14.088 ml = 32.470 ml