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.
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?
1 answer