Question
What is the pH of a 100 mL solution of 100 mM acetic acid at pH 3.2 following addition of 5 mL of 1 M NaOH? The pKa of acetic acid is 4.70.
The answer is suppose to be 4.75. It will nice to show the steps on how to get this answer. Thank you
But I keep getting 5.70 instead!?
The answer is suppose to be 4.75. It will nice to show the steps on how to get this answer. Thank you
But I keep getting 5.70 instead!?
Answers
I'm confused with the numbers. The pH of a 100 mL acetic acid solution that is 0.1 M (100 mM) is 1.4E-3 for (H^+) or pH of 2.85 and not 3.20. Does this mean the solution is not pure acetic acid?
Also did you make a typo on pKa acetic acid. Is that really 4.75 and no 4.70.
Assuming it is a pure acetic acid solution and that pKa is 4.75, which is what I think you have, I would do this. HAc is acetic acid.
millimols HAc = mL x M = 100 x 0.1 = 10
millimols NaOH added = 5 x 1 = 5
........... HAc + OH ==> Ac^- + H2O
I..............10.........0...........0...........0
add......................5..............................
C............-5..........-5............+5...............
E.............5............0..............5
Plug into the Henderson-Hasselbalch equation like this.
pH = pKa + log (base)/(acid)
pH = 4.75 + log (5/5) = 4.75 + 0 = 4.75
Don't hesitate to correct me if I have made the wrong assumptions.
Also did you make a typo on pKa acetic acid. Is that really 4.75 and no 4.70.
Assuming it is a pure acetic acid solution and that pKa is 4.75, which is what I think you have, I would do this. HAc is acetic acid.
millimols HAc = mL x M = 100 x 0.1 = 10
millimols NaOH added = 5 x 1 = 5
........... HAc + OH ==> Ac^- + H2O
I..............10.........0...........0...........0
add......................5..............................
C............-5..........-5............+5...............
E.............5............0..............5
Plug into the Henderson-Hasselbalch equation like this.
pH = pKa + log (base)/(acid)
pH = 4.75 + log (5/5) = 4.75 + 0 = 4.75
Don't hesitate to correct me if I have made the wrong assumptions.
The pka=4.70
And the answer for new ph= 4.75
And the answer for new ph= 4.75
If you had answered my questions this would have been easier. It is NOT possible to have a 0.1 M pure acetic acid solution with a pH of 3.2; therefore, it MUST be a buffer and that probably is your basic problem. Here is how you do it.
pH = pKa + log (base/acid)
3.2 = 4.70 + log b/a. You work it out and
b/a 0.0316 or base = 0.0316*acid
100 mL of the 0.1 M buffer is 10 millimoles. therefore,solve these two equation simultaneously.
acid + base = 10
acid + 0.0315*acid = 10 and
acid = 10/1.0316 = 9.69 and base = 0.306. That is what's in the 0.1 M buffer that you called a 0.1 M acetic acid solution. Now set up an ICE this way but I will call acetic acid HAc(acid) and acetate is Ac^-(base)
You add 5 mL x 1 M NaOH = 5 millimoles NaOH
...............HAc + OH^- ==> Ac^- + H2O
I...............9.69......0..............0.316
add......................5...........................
C...............-5........-5..............+5
E.............4.69........0..............5.316
pH = pKa + log base/acid
pH = 4.70 + log (5.316/4,69).
Solve and pH = 4.75
pH = pKa + log (base/acid)
3.2 = 4.70 + log b/a. You work it out and
b/a 0.0316 or base = 0.0316*acid
100 mL of the 0.1 M buffer is 10 millimoles. therefore,solve these two equation simultaneously.
acid + base = 10
acid + 0.0315*acid = 10 and
acid = 10/1.0316 = 9.69 and base = 0.306. That is what's in the 0.1 M buffer that you called a 0.1 M acetic acid solution. Now set up an ICE this way but I will call acetic acid HAc(acid) and acetate is Ac^-(base)
You add 5 mL x 1 M NaOH = 5 millimoles NaOH
...............HAc + OH^- ==> Ac^- + H2O
I...............9.69......0..............0.316
add......................5...........................
C...............-5........-5..............+5
E.............4.69........0..............5.316
pH = pKa + log base/acid
pH = 4.70 + log (5.316/4,69).
Solve and pH = 4.75
Thank you for your help! @DrBob222