What is the pH before equivalence, after equivalence, and at equivalence?

Question

A chemist titrates 90.0mL of a 0.8055M dimethylamine CH32NH solution with 0.6459M HNO3 solution at 25°C. Calculate the pH at equivalence. The pKb of dimethylamine is 3.27.Round your answer to 2 decimal places.

Answer 1

First, let's define what we mean by equivalence. The equivalence point is when the moles of base (CH₃)₂NH, and acid, HNO3 are present in stoichiometric amounts.

(CH₃)₂NH + HNO₃ == (CH₃)₂NH₂⁺ + NO₃^- ... balanced equation

In the current problem we have ...

mols (CH₃)₂NH = 90.0 mls x 1 L / 1000 mls x 0.8055 mol / L = 0.072495 mols (CH₃)₂NH

At the equivalence point, you would have 0.072495 mols of HNO3 since they are 1:1 in stoichiometry

0.072495 mol HNO₃ x 1 L / 0.6459 mols = 0.1122 L = 112.2 mls

Total volume of reaction at the equivalence point = 90.0 ml + 112.2 mls = 202.2 mls = 0.2022 L

At this point, all of the (CH₃)₂NH has been converted to (CH₃)₂NH₂⁺ (which is the conjugate acid)

[(CH₃)₂NH₂⁺] = 0.072459 mols / 0.2022 L = 0.3584 M

To find pH, we must look at hydrolysis of (CH₃)₂NH₂⁺

(CH₃)₂NH₂⁺ + H₂O ==> H₃O⁺ + (CH₃)₂NH

Ka = [H₃O⁺][(CH₃)₂NH] / [(CH₃)₂NH₂⁺]

From pKb we can find Ka: pKb + pKa = 14 thus pKa = 10.73 and Ka = 1x10-^10.73 = 1.86x10^-11

1.86x10^-11 = (x)(x) / 0.3584-x (assume x is small and ignore it in denominator)

x² = 6.67x10^-12

x = 2.58x10^-6 = [H₃O⁺]

pH = -log 2.58x10^-6

pH = 5.59

The pH before equivalence and after equivalence will depend on how much before and after the equivalence point you are. Without knowing the moles of HNO3 added, we can't calculate the actual pH.