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
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.
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.
To calculate the resulting pH after the addition of acid or base to a buffer solution, we need to consider the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where:
- pH is the resulting pH of the buffer solution
- pKa is the logarithmic acid dissociation constant of the weak acid
- [A-] is the concentration of the conjugate base
- [HA] is the concentration of the weak acid
First, let's calculate the pH after the addition of 2 mL of 0.1 M HCl to 10 mL of the 8:2 buffer (acetic acid to sodium acetate).
Given:
- [CH3COOH] = 0.1 M
- [CH3COONa] = 0.1 M
- Ratio = 8:2
1. Calculate the concentrations of acetic acid and sodium acetate in the buffer solution:
- Acetic acid (CH3COOH): 10 mL * (8 / (8+2)) = 8 mL
[CH3COOH] = (8 mL / 10 mL) * 0.1 M = 0.08 M
- Sodium acetate (CH3COONa): 10 mL * (2 / (8+2)) = 2 mL
[CH3COONa] = (2 mL / 10 mL) * 0.1 M = 0.02 M
2. Calculate the initial concentration of CH3COO-:
- [CH3COO-] = 0.02 M
3. Calculate the initial concentration of CH3COOH:
- [CH3COOH] = 0.08 M
4. Calculate the pKa of acetic acid:
The pKa value for acetic acid is around 4.75.
5. Calculate the resulting pH after the addition of 2 mL of 0.1 M HCl:
- Assuming the 2 mL of HCl added reacts completely, the new concentration of CH3COOH would be:
[CH3COOH]new = [CH3COOH]initial - (volume of HCl added * concentration of HCl) = 0.08 M - (2 mL * 0.1 M)
- Assuming the volume does not change significantly, the new concentration of CH3COONa would be:
[CH3COONa]new ≈ [CH3COONa]initial = 0.02 M
- Using the Henderson-Hasselbalch equation:
pH = pKa + log([CH3COONa]new / [CH3COOH]new)
Substitute the values:
pH = 4.75 + log(0.02 M / 0.06 M)
Now, let's calculate the resulting pH after the addition of 4 mL of 0.1 M NaOH to 10 mL of the 8:2 buffer (acetic acid to sodium acetate).
Given:
- [CH3COOH] = 0.1 M
- [CH3COONa] = 0.1 M
- Ratio = 8:2
1. Calculate the concentrations of acetic acid and sodium acetate in the buffer solution:
- Acetic acid (CH3COOH): 10 mL * (8 / (8+2)) = 8 mL
[CH3COOH] = (8 mL / 10 mL) * 0.1 M = 0.08 M
- Sodium acetate (CH3COONa): 10 mL * (2 / (8+2)) = 2 mL
[CH3COONa] = (2 mL / 10 mL) * 0.1 M = 0.02 M
2. Calculate the initial concentration of CH3COO-:
- [CH3COO-] = 0.02 M
3. Calculate the initial concentration of CH3COOH:
- [CH3COOH] = 0.08 M
4. Calculate the pKa of acetic acid:
The pKa value for acetic acid is around 4.75.
5. Calculate the resulting pH after the addition of 4 mL of 0.1 M NaOH:
- Assuming the 4 mL of NaOH added reacts completely, the new concentration of CH3COONa would be:
[CH3COONa]new = [CH3COONa]initial + (volume of NaOH added * concentration of NaOH) = 0.02 M + (4 mL * 0.1 M)
- Assuming the volume does not change significantly, the new concentration of CH3COOH would be:
[CH3COOH]new ≈ [CH3COOH]initial = 0.08 M
- Using the Henderson-Hasselbalch equation:
pH = pKa + log([CH3COONa]new / [CH3COOH]new)
Substitute the values:
pH = 4.75 + log(0.06 M / 0.08 M)
Now, you can calculate the resulting pH for both cases by substituting the values into the Henderson-Hasselbalch equation.