How do i calculate the pH when 200ml of .200M HCl is added .4998M of sodium acetate?
3 answers
This problem reads suspiciously like something is missing. ......of 0.200 M HCl is added something here .4998 M socium acetate. I don't know how much sodium acetate you have.
sorry huge typo
How do i calculate the pH when 200ml of .200M HCl is added to 200.0 ml of .4998M of sodium acetate?
How do i calculate the pH when 200ml of .200M HCl is added to 200.0 ml of .4998M of sodium acetate?
THANK YOU. I've been looking at this thing, of and on, for over an hour. Even tried some dry runs to see it might work out if I simply assumed a volume of sodium acetate. Of course, none of them would pan out because it made a difference in how much I assumed. SO,
let's call sodium acetate NaAc and the acetate ion Ac^-. Helps on typing.
Ac^- + H^+ ==> HAc (acetic acid)
initial:
Ac^- = M x mL = 0.4998 x 200 mL = 99.96 millimoles.
H^+ = M x mL = 0.2 x 200 mL = 40 mmoles.
HAc = 0 mmoles.
final:
HAc = 40 mmoles.
H^+ = 0
Ac^- = 99.96-40 = 59.96
(Ac^-) = mmols/mL (or moles/L)
(HAc) = mmols/mL (or moles/L)
Substitute into HH equation. Base is Ac^- and acid is HAc.
I get something like 4.9 or so but that's just a close estimate. You need to do it more accurately. By the way, the mL in the concn conversion is 400 mL.
let's call sodium acetate NaAc and the acetate ion Ac^-. Helps on typing.
Ac^- + H^+ ==> HAc (acetic acid)
initial:
Ac^- = M x mL = 0.4998 x 200 mL = 99.96 millimoles.
H^+ = M x mL = 0.2 x 200 mL = 40 mmoles.
HAc = 0 mmoles.
final:
HAc = 40 mmoles.
H^+ = 0
Ac^- = 99.96-40 = 59.96
(Ac^-) = mmols/mL (or moles/L)
(HAc) = mmols/mL (or moles/L)
Substitute into HH equation. Base is Ac^- and acid is HAc.
I get something like 4.9 or so but that's just a close estimate. You need to do it more accurately. By the way, the mL in the concn conversion is 400 mL.
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