In: Chemistry
Calculate the pH of the solution after the addition of the following amounts of 0.0615 M HNO3 to a 50.0 mL solution of 0.0750 M aziridine. The pKa of aziridinium is 8.04.
a) 0.00 mL of HNO3
b) 9.69 mL of HNO3
c) Volume of HNO3 equal to half the equivalence point volume
d) 57.1 mL of HNO3
e) Volume of HNO3 equal to the equivalence point
f) 65.3 mL of HNO3
Solution:
Given data: pKa = 8.04 and 0.0750 M aziridine
The reaction follows, HNO3 + Aziridine -----> HA + NO3-
Keq = Ka/Kb = [HA][NO3-] / [HNO3][Aziridine]
Kb = [HNO3][Aziridine]
= (0.0615)(0.0750) = 0.0046125
pKb = -log (0.0046125) = 2.336063 = 2.336
By using Henderson- Hasselbalch equation pH can be calculated using known pK values,
pKa = pH - log{[Salt] / [Acid]} here aziridinium is salt and HNO3 is acid
8.04 = pH - log {0.0750/0.0615} = pH - 0.0860
pH = 8.04 + 0.0860 = 8.126
Part (a)
pOH = 0.05(pKb - lg(0.075)) = 4.1
14 = pH + pOH = pH + 4.1
pH = 9.9
Part(b)
at 9.69 mL Molarity [M] = {0.00969/63}/0.05 = 0.003076 M
pKa = pH - log {[Salt] / [Acid]}
8.04 = pH - log {[0.0750] / [0.003076]}
8.04 = pH - 1.3870
pH = 9.427
Part(c)
At this point pH = pKa = 8.04
Part(d)
at 57.1 mL Molarity [M] = {0.0571/63}/0.05 = 0.001812 M
pKa = pH - log {[Salt] / [Acid]}
8.04 = pH - log {[0.0750] / [0.001812]}
8.04 = pH - 1.6169
pH = 9.656
Part (e)
at the equivalence point {57.1+65.3} /2 = 61.2 mL
at 61.2 mL Molarity [M] = {0.0612/63}/0.05 = 0.001942 M
pKa = pH - log {[Salt] / [Acid]}
8.04 = pH - log {[0.0750] / [0.001942]}
8.04 = pH - 1.5868
pH = 9.6268
Part (f)
in excess of 65.3 - 61.2 = 4.1 mL
at excess 4.1 mL Molarity [M] = {0.0041/63}/0.05 = 0.001301 M
pKa = pH - log {[Salt] / [Acid]}
8.04 = pH - log {[0.0750] / [0.001301]}
8.04 = pH - 1.7607
pH = 9.8007