Asked by Elaine
a 25.0 mL sample of .050 M solution of aqueous trimethylamine is titrated with a .063 M solution of HCl. calculate pH the solution after 10.0 mL, 20.0 mL, 30.0 mL of acid have been added. pKb of (CH3)_3_N= 4.19 at 25 degrees C
Answers
Answered by
DrBob222
First determine where the equivalence point is so you will know where you are on the titration curve with each of thes additions.
mL x M = mL x M
25.0 x 0.05 = mL x 0.063
mL = about 19 or so.
Therefore, 10 mL will be before the e.p. and you will have some base and some of its conjugate. Use the Henderson-Hasselbalch equation.
b. 20 mL will be just after the e.p. So you will have an excess of 20 mL-mL for e.p. That will be an excess of HCl
c. 30 mL is same as part b.
mL x M = mL x M
25.0 x 0.05 = mL x 0.063
mL = about 19 or so.
Therefore, 10 mL will be before the e.p. and you will have some base and some of its conjugate. Use the Henderson-Hasselbalch equation.
b. 20 mL will be just after the e.p. So you will have an excess of 20 mL-mL for e.p. That will be an excess of HCl
c. 30 mL is same as part b.
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.