No, you didn't substitute correctly. The 20 you have should be 20-x
5.2 = 4.75 + log(x/20-x) and solve for x.
You last step shown should be
56.37 = x/(20-x)
Also, I don't know where you obtained 1.75.
x should be approximately 15 mmoles and at 1 M NaOH that would be approximately 15 mL.
If you have 200 ml of a 0.1M solution of acetic acid (pka=4.75),
1. How many ml of a solution of 1.O M NaOH would you need to adjust the pH to 5.2?
"Chemistry - DrBob222, Wednesday, June 29, 2011 at 8:41pm
200 mL x 0.1M = 20 mmoles.
1.
............HAc + OH^- ==> Ac^- H2O
initial.....20.0...0........0
added..............x...........
change.......-x....-x........+x
equil......20-x.....0.........+x
Substitute into the Henderson-Hasselbalch equation and solve for x which will be mmoles NaOH. You can take it from there."
So I have substituted for hernderson hasselbalch
5.2= 4.75 + {log [x]/[20 mmoles]
1.75= log [x]/20mmoles
10^1.75= [x]/20
56. 234 = x/20
1,124.68 mmoles of NaOH.
Could this be right? Now to convert this to ml to answer the question, would I divided by 1.0M? So I would need 1.124 L ?
3 answers
I dont' know what I was thinking with the 1.75. So the 20-x indicates the amount of acid left over after the base reacts with it. Thanks a lot Dr. Bob, I really appreciate it.
That's right. You want to substitute the EQUILIBRIUM numbers and not the STARTING numbers.