Asked by lilly
What is the pH of a buffer solution if you have 250 ml of a 1.56M Acetic Acid and you added 26.56 grams of sodium acetate (NaCH3CO2)? What is the new pH if you now add 1gram of NaOH to the buffer solution?
Answers
Answered by
DrBob222
A.
pH = pKa + log (base)/(acid)
Convert 26.56 g NaAc to mols, convert to M, plug into the above equation and solve for pH.
B.
I like to work these by millimols (mmol) and not M.
mmols HAc = 250 x 1.56 = about 400 but you need a more accurate number.
mmols NaAc = 26.56/82 x 1000 = about 320--again you need a more accurate answer.
mmols NaOH = 1/40 x 1000 = about 25.
.........HAc + OH^- ==> Ac^- + H2O
I........400...0........320.......
add...........25...............
C........-25..-25.......+25
E........375...0........345
Substitute the E line into the HH equation and solve for the new pH. Remember those numbers above are approximations; you need to redo the entire problem using the correct numbers. Post your work if you get stuck.
pH = pKa + log (base)/(acid)
Convert 26.56 g NaAc to mols, convert to M, plug into the above equation and solve for pH.
B.
I like to work these by millimols (mmol) and not M.
mmols HAc = 250 x 1.56 = about 400 but you need a more accurate number.
mmols NaAc = 26.56/82 x 1000 = about 320--again you need a more accurate answer.
mmols NaOH = 1/40 x 1000 = about 25.
.........HAc + OH^- ==> Ac^- + H2O
I........400...0........320.......
add...........25...............
C........-25..-25.......+25
E........375...0........345
Substitute the E line into the HH equation and solve for the new pH. Remember those numbers above are approximations; you need to redo the entire problem using the correct numbers. Post your work if you get stuck.
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