Asked by Chemistry

How many mL of .200M acetic acid are mixed with 13.2mL of.200M sodium acetqte to give a buffer with pH=4.2
Answered by DrBob222
Use the Henderson-Hasselbalch equation.
Answered by Chemistry
I'm making a mistake in using that
equation:

n (CH3COO-)= .0132*.2 = .oo264

..does that mean that CH3COOH also has the same n?

ph= pKa +log(Conjugate Base/ Weak Acid)
4.2= pKa + log (.00264/ .00264)

since i have n the same it doesn't work. How do you calculate n?
Answered by DrBob222
pKa = 4.76 are thereabout. Look it up if you know it.
4.2 = 4.76 + log (base/acid)
-0.56 = log (B/A)
-0.56 = log (0.00264/A)
0.275 = 0.00264/A
So the acid (acetic acid) must be
0.00264/0.275 = 0.00960 moles (technically molar but the volume always cancels and you can ingore that for the moment.)
M x L = moles to find the L necessary.
Check my arithmetic. My calculator rounded here and there so you need to go through it again to make sure.
Answered by Chemistry
how did you calculate pKa because this is a practise question for my test and we won't be given the value. so how would you calculate it?
Answered by DrBob222
Don't you have Ka for acetic acid = 1.75 x 10^-5 or something like that.
Then pKa = -log Ka = -log 1.75 x 10^-5 = -(-4.75696) = 4.75696 which I rounded to 4.76. And I get something like 48 mL of the acetic acid required but check my math.
Answered by Chemistry
Neither Ka nor Pka was given
Answered by DrBob222
On tests I have seen, they usually give Ka or pKa in the problem OR there will be an appended set of tables that will contain constants necessary to take the exam. I don't know of any test in which you are expected to know those constants.
Answered by Chemistry
okie thank you =)
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