Can you calculate the pH before equivalence, after equivalence, and at equivalence?

Question

An analytical chemist is titrating 50.0mL of a 0.5100M solution of dimethylamine CH32NH with a 0.4000M

 solution of HNO3. The pKb of dimethylamine is 3.27. Calculate the pH of the base solution after the chemist has added 73.4mL of the HNO3 solution to it.

Answer 1

In a previous answer, I gave you the calculation for pH at the equivalence point. Below is the calculation for pH after the equivalence point when you have added 73.4 mls of 0.4000 M HNO3. Cannot find pH before equivalence point unless we know the volume of HNO3 added at that point.

(CH₃)₂NH + HNO₃ ==> (CH₃)₂NH₂⁺ + NO₃^-

moles (CH₃)₂NH initially present = 50.0 ml x 1 L / 1000 ml x 0.5100 mol / L = 0.02550 moles

moles HNO3 added = 73.4 ml x 1 L / 1000 ml x 0.4000 mol / L = 0.02936 mols

(CH₃)₂NH + H⁺ ==> (CH₃)₂NH₂⁺

0.0255......0.02936...........0...............Initial

-00255....-0.0255..........+0.0255.......Change

0............0.00386...........0.0255........Equilibrium

So, after addition of 73.4 ml of HNO₃, we have an excess of H⁺ in the amount of 0.00386 moles.

Final volume = 50.0 ml + 73.4 ml = 123.4 mls = 0.1234 L

[H⁺] = 0.00386 mols / 0.1234 L = 0.03128 mol/L

pH = -log [H⁺] = -log 0,03128

pH = 1.50

(note: this calculation ignores the contribution of (CH₃)₂NH₂⁺ to the pH as the Ka is very low)