How many mL of .200M acetic acid are mixed with 13.2mL of.200M sodium acetqte to give a buffer with pH=4.2

Answer 1

Use the Henderson-Hasselbalch equation.

Answer 2

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?

Answer 3

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.

Answer 4

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?

Answer 5

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.

Answer 6

Neither Ka nor Pka was given

Answer 7

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.

Answer 8

okie thank you =)