posted by Chemistry on .
How many mL of .200M acetic acid are mixed with 13.2mL of.200M sodium acetqte to give a buffer with pH=4.2
Use the Henderson-Hasselbalch equation.
I'm making a mistake in using that
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?
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.
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?
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.
Neither Ka nor Pka was given
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.
okie thank you =)