Asked by ALISON
a 250.0ml buffer solution is 0.250M in ascetic acid and 0.250M in sodium acetate.
B). What is pH after addition of 0.0050mol of HCl?
C). What is pH after addition of 0.0050mol of NaOH?
B). What is pH after addition of 0.0050mol of HCl?
C). What is pH after addition of 0.0050mol of NaOH?
Answers
Answered by
DrBob222
250 mL x 0.250M = 62.5 millimols HAc
250 mL x 0.250M = 62.5 mmols NaAc.
add 5 mmols HAc
add 5 mmols NaOH
............Ac^- + H^+ ==> HAc
initial.....62.5..........62.5
add................5...........
change......-5....-5......+5
equil.....57.5.....0......67.5
Substitute into the Henderson-Hasselbalch equation and solve for pH.
The same process will work the NaOH part.
equil.....
250 mL x 0.250M = 62.5 mmols NaAc.
add 5 mmols HAc
add 5 mmols NaOH
............Ac^- + H^+ ==> HAc
initial.....62.5..........62.5
add................5...........
change......-5....-5......+5
equil.....57.5.....0......67.5
Substitute into the Henderson-Hasselbalch equation and solve for pH.
The same process will work the NaOH part.
equil.....
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