PKa? Ka? Determination of pK, Values of Unknown Acid First determination pH Seco
ID: 1044187 • Letter: P
Question
PKa? Ka? Determination of pK, Values of Unknown Acid First determination pH Second determination mL NaOH ?.OO Third determination pH mL NaOH pH mL NoOH 2.o0 3.0 2.0? ?. a.20 .05 a.3o .$9 2:43 7?? a.62 8. ?? 109193a 5. ?? 2.O 12.93 3.05 2.03 13.0 53 .97 I.O3 lt.oo 17.03 ??.os 18.42 II33 10.87 19001.to o.7 15.03 II.3 ?.7g I. 49 19.03 1.76 I 78 22.03 23.0? 22.0? 22.0 12.?? Copyright O2013 Pearson Education, Inc. pk pKo Kal Kal Kal Standard deviation Average Kat- PK Kaz Standard deviation Average Ka Copyright C 2015 Pearson Education, Ine
Explanation / Answer
From
First determination
pKa = -log[Ka]
pKa1 = 2.04
Ka = 9.12 x 10^-3
pKa2 = 5.83
Ka2 = 1.48 x 10^-6
Second determination
pKa1 = 2.25
Ka1 = 5.62 x 10^-3
pKa2 = 5.43
Ka2 = 3.71 x 10^-6
Third determination
pKa1 = 2.06
Ka1 = 8.71 x 10^-3
pKa2 = 5.53
Ka2 = 2.95 x 10^-6
average Ka1 = 7.82 x 10^-3
Standard deviation = sq.rt.[sum(x-mean]^2/2]
= sq.rt.[(1.69 x 10^-6 + 4.84 x 10^-6 + 7.92 x 10^-7)/2]
= 1.91 x 10^-3 [Ka1]
Standard deviation = sq.rt.[sum(x-mean]^2/2]
= sq.rt.[(1.513 x 10^-12 + 1 x 10^-12 + 5.76 x 10^-14)/2]
= 1.13 x 10^-6 [Ka2]
average Ka2 = 2.71 x 10^-6
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