Calculate the pH of the solution after the addition of the following amounts of

Question

Calculate the pH of the solution after the addition of the following amounts of 0.0545 M HNO3 to a 70.0 mL solution of 0.0750 M aziridine. The pKa of aziridinium is 8.04.

e) volume of HNO3 equal to the equivalence point

pH =

answer is not: 8.04, 3.27, or 4.23

hint given:

Calculate the concentration of azaridinium by dividing the original moles of azaridine in solution by the new total volume. Solve for the equilibrium concentration of H3O as you would in a typical weak acid problem by setting up an ICE table and using the Ka expression to determine how much aziridinium dissociates into azaridine at equilibrium. The H3O concentration is then converted to pH. 4) After the equivalence point, the solution contains a mixture of the weak acid and excess strong acid (HNO3). The pH is determined from the concentration of the excess strong acid, while the contribution of the weak acid is neglected. Determine the concentration of the excess H in solution. Calculate the moles of H added to the solution and subtract the original moles of aziridine in solution. Divide the excess moles of H by the new total volume. The H concentration is then converted to pH.

Explanation / Answer

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