Asked by Katie
Calculate the pH at the equivalence point in the titration of 50ml of 0.20 M methylamine (Kb=4.3*10^-4) with a 0.40 M HCL solution.
Answers
Answered by
DrBob222
How many mL HCl is needed? That's
50 x 0.2/0.4 = 25.00 mL
Total volume = 75.00 mL
Total mols = 0.050 x 0.2 =0.01
(salt at equivalence point) = 0.01/0.075L = approximately 0.13
If we call methylamine just BNH2 then the salt will be BNH3^+ and it will hydrolyze as follows:
......BNH3^+ + H2O ==> H3O^+ + BNH2
I.....0.13..............0........0
C.......-x..............x........x
E...-0.13-x.............x........x
Ka for BNH3^+ = (Kw/Kb for BNH2) = (x)(x)/(0.13-x)
Solve for x = (H3O^+) and convert to pH.
50 x 0.2/0.4 = 25.00 mL
Total volume = 75.00 mL
Total mols = 0.050 x 0.2 =0.01
(salt at equivalence point) = 0.01/0.075L = approximately 0.13
If we call methylamine just BNH2 then the salt will be BNH3^+ and it will hydrolyze as follows:
......BNH3^+ + H2O ==> H3O^+ + BNH2
I.....0.13..............0........0
C.......-x..............x........x
E...-0.13-x.............x........x
Ka for BNH3^+ = (Kw/Kb for BNH2) = (x)(x)/(0.13-x)
Solve for x = (H3O^+) and convert to pH.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.