2) Rank the following titrations in order of increasing pH at the equivalence po
ID: 563859 • Letter: 2
Question
2) Rank the following titrations in order of increasing pH at the equivalence point of the titration (1 = lowest pH and 5 = highest pH).
__ 200.0 mL of 0.100 M HC2H3O2 (Ka = 1.8 x 10-5) by 0.100 M NaOH
__ 100.0 mL of 0.100 M HNO2 (Ka = 4.0 x 10-4) by 0.100 M NaOH
__ 200.0 mL of 0.100 M (C2H5)2NH (Kb = 1.3 x 10-3) by 0.100 M HCl
__ 100.0 mL of 0.100 M KOH by 0.100 M HCl
__ 200.0 mL of 0.100 M H2NNH2 (Kb = 3.0 x 10-6) by 0.100 M HCl
Explanation / Answer
200.0 mL of 0.100 M HC2H3O2 (Ka = 1.8 x 10-5) by 0.100 M NaOH
At equivalence point,
PH = 0.5 * PKw + 0.5 * PKa + 0.5 log C
PKa = -log (1.8 * 10^-5) = 4.74
PH > 7
100.0 mL of 0.100 M HNO2 (Ka = 4.0 x 10-4) by 0.100 M NaOH:
PKa = - log (4.0 - 10^-4) = 3.4
higher the PKa means higher the PH.
here PH > 7
__ 200.0 mL of 0.100 M (C2H5)2NH (Kb = 1.3 x 10-3) by 0.100 M HCl
PKb = - log Kb = - log (1.3 * 10^-3) = 2.89
__ 100.0 mL of 0.100 M KOH by 0.100 M HCl: as NaOH is strong base and HCl is strong acid. so PH at equivalence point is 7.0.
__ 200.0 mL of 0.100 M H2NNH2 (Kb = 3.0 x 10-6) by 0.100 M HCl:
At equivalence point salt hydrolysis occurs.
PH = 0.5 * PKw - 0.5 * PKb - 0.5 * log C
PKb = - log (3.0 * 10^6) = 5.52
higher PKb value means lower the PH.
so 1 = lowest pH and 5 = highest pH:
5) 200.0 mL of 0.100 M HC2H3O2 (Ka = 1.8 x 10-5) by 0.100 M NaOH
4) 100.0 mL of 0.100 M HNO2 (Ka = 4.0 x 10-4) by 0.100 M NaOH
3) 100.0 mL of 0.100 M KOH by 0.100 M HCl.
2) 200.0 mL of 0.100 M (C2H5)2NH (Kb = 1.3 x 10-3) by 0.100 M HCl.
1) 200.0 mL of 0.100 M H2NNH2 (Kb = 3.0 x 10-6) by 0.100 M HCl.
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