1. based on the curve shown, what is the value of the equivalence point? 2. how
ID: 747597 • Letter: 1
Question
1. based on the curve shown, what is the value of the equivalence point?
2. how many ml did it require to reach this value?
3. What is the pKa of the acid?
4. Is this an SB/SA, WA, or WB/WA titration curve?
Explanation / Answer
find the molarity of the 65.0 ml of acid solution: since they react in a 1mole:1mole ratio we can use the simplified titration formula (millimoles = millimoles): M1V1 = M2V2 (Molarity of acid) (65.0ml) = (0.06209 M NaOH)(44.18ml) Molarity of acid: 0.04220 M fond the total molesof acid in 65.0 ml: 0.0650 ml @ 0.04220 mol/litre = 0.002743 moles acid total find the moles of NaOH added in 29.64ml , when the pH was 3.67: 0.02964 litres @ 0.06209 mol/litre = 0.001840 moles NaOH added @ 29.64 mls, when: 1 NaOH & 1 HX => NaX & H2O adding 0.001840 moles NaOH reacts with 0.001840 moles acid: 0.002743 mol acid total - 0.001840 mol reacts = 0.000903 mol remain @ 29.64 mls, when: 1 NaOH & 1 HX => NaX & H2O adding 0.001840 moles NaOH reacts with 0.001840 moles acid: it produces 0.001840 moles of NaX @ 29.64 mls, find the new molarities that are in the 94.65 total mls: 0.000903 mol acid / 0.09465 litres = 0.00954 M [HX acid] 0.001840 mol NaX / 0.09465 litres = 0.01944 M [X- ion] pH of 3.67 , gives a [H+] = 2.138 e-4 M Ka = [H+] [X-] / [HX] Ka = [2.138 e-4] [0.01944] / [0.00954] Ka = 4.357e-4 your answer(2sigfigs): K = 4.4e-4 (the pH of 3.67, had only 2 sigfigs)
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