**Unformatted text preview: **Using the graph, calculate the apparent concentra±on of this tripro±c acid and the dissocia±on constants for all proton dissocia±on steps (K a1 , K a2 , and K a3 ). Approxima±ons of points on the graph 1 st equivalence point (12.5 ml, 4.9 pH) 2 nd equivalence point (25.5 mL, 8.1 pH) Midpoint 1 (6.0 mL, 3.62 pH) Midpoint 2 (18.5 mL, 6.2 pH) Midpoint 3 (31.25 mL, 9.5 pH) pKa1 = 3.62 pKa2 = 6.2 pKa3 = 9.5 K a1 = 10 3.62 = 2.40 10 4 K a2 = 10 6.2 = 6.3 10 7 K a3 = 10 9.5 = 3.2 10 10 It took 12.5 mL to reach ³rst equivalence point and 13 mL (25.5 mL – 12.5 mL) to reach second equivalence point. Average = 12.75 mL per equivalence point ((12.5 mL + 13 mL)/2)) 0.0800 mol/L NaOH x 0.01275 L NaOH = 0.00102 mol NaOH Moles of tripro±c acid = moles of NaOH Therefore there are 0.00102 mol tripro±c acid 0.00102 mol tripro±c acid/0.07500 L tripro±c acid = 0.0136 M tripro±c acid The concentra±on of this tripro±c acid is 0.0136 M....

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- Spring '08
- LEMASTER
- Chemistry, Valence, pH, Equivalence point, Triprotic Acid