Calculate the pH of the solution after the addition of each of
the given amounts of 0.0637 M HNO30.0637 M HNO3 to a 60.0 mL60.0 mL
solution of 0.0750 M0.0750 M aziridine. The p?apKa of aziridinium
is 8.04.
What is the pH of the solution after the addition of 74.9" mL
HNO3?
60.0 mL solution of 0.0750 M aziridine = 0.060 L * 0.0750 mole / L = 0.0045 mole.
and
74.9" mL of 0.0637 M HNO3 = 0.0749 L * 0.0637 mole / L = 0.00477 mole.
excess HNO3 = (0.00477 - 0.0045) = 2.71 * 10^-4 mole.
total volume = (60.0 + 74.9) = 134.9 ml = 0.1349 L
molarity of excess HNO3 = 2.71 * 10^-4 mole / 0.1349 L = 2.01 * 10^-3 M
pH = - log [H+]
or
PH = - log (2.01 * 10^-3 )
or
PH = 2.70
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