I will call CH3COONa = NaAc. I don't see a question about the 2:8 ratio; therefore, I will limit my response to the 8:2 ratio. If we assume we have 80 mL HAc and 20 mL of NaAc we have the following:
millimoles HAc = 80 mL x 0.1 M = 8
millimoles NaAc = 20 mL x 0.1 M = 2
If we take 10 mL of the buffer then we have 8 x (10 mL/100 mL) = 0.8 HAc
and 2 x (10/100) = 0.2 millimoles NaAc.
adding 2 mL x 0.1 M HCl = 0.2 millimoles HCl
.........................Ac^- + HCl ==> HAc + Cl^-
I........................0.2........0...........0.8.........0
add.............................0.2...........................
C.....................-0.2.....-0.2.............+0.2.................
E......................0...........0................1.0
So the final solution is zero Ac, zero HC, and 1.0 mmoles HAc in a volume of 10 mL buffer + 2 mL added HCl = 12 mL total. (HAc) in the final solution is M = mmols/mL = 1/12 = 0.0833
I will leave that for you to do. You have a 0.0833 M CH3COOH and you want to determine the pH.
Now for the added NaOH to the 10 mL of the buffer.
millimoles NaOH added = mL x M = 4 mL x 0.1 M = 0.4 mmoles.
...................HAc + OH^- ==> Ac^- + H2O
I...................0.8........0.............0.2......0
add.........................0.4...................................
C................-0.4.....-0.4...........+0.4.....................
E.................0.4.........0.............0.6..........................
So the final solution is 0.4 mmoles HAc + 0.6 mmoles NaAc. Plug these into the Henderson-Hasselbalch equation of pH = pKa + log [(acetate)/acid)]
and solve for pH.
Post your work if you get stuck.
The acetate buffer were prepared by mixing 0.1 M ch3cooh with 0.1M ch3cooNa in the ratio 8:2 and 2:8. Calculate the resulting ph after the addition of 2mL of 0.1 M HCl to 10 ml of buffer 8:2 and 4ml of 0.1 M NaOH to 10 ml of buffer 8:2
2 answers
Let me point out on the H-H equation that it is
pH = pKa + log [(base)/(acid)]. Technically, that is concentration is moles/L and I did not use mols/L or millimoles/mL. I used millimoles ONLY; however, that is OK because if (base) = millimoles/mL and (acid) = millimoles/mL, this is ONE solution, the volume is the same so mL in the numerator cancels with mL in the denominator and millimoles/millimoles gives the same numerical answer.
pH = pKa + log [(base)/(acid)]. Technically, that is concentration is moles/L and I did not use mols/L or millimoles/mL. I used millimoles ONLY; however, that is OK because if (base) = millimoles/mL and (acid) = millimoles/mL, this is ONE solution, the volume is the same so mL in the numerator cancels with mL in the denominator and millimoles/millimoles gives the same numerical answer.