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A biochemist prepares a lactate buffer by mixing 246.5 mL of 0.79 M lactic acid (Ka = 1.38×10-4) with 435.8 mL of 0.62 M sodium lactate. What is the pH of the buffer?

I cant seem to get this question. I set up my ICE Table and got an answer for the H+ of H30. Help please..

2007-03-14 14:32:29 · 3 answers · asked by Anonymous in Science & Mathematics Chemistry

3 answers

For buffers you can get away with using the Henderson-Hasselbalch equation (and assumptions)

pH=pKa+log[con.base]/[acid] =
=pKa +log( mmole conj.base/ mmole acid )
since V and the conversion factor from mole to mmole are simplified in the ratio
I use mmole so that I don't have to convert mL to L

pH=(-log(1.38*10^-4) ) + log (0.62*435.8/(0.79*246.5)) = 4.00

It is the same as setting up the ICE and doing the assumption that x<< Cacid, Conj.base so that the initial concentrations just after mixing are practically equal to the concentrations at equilibrium.

With this approximation you will reach the same result as with the exact solution (solving the quadratic that comes from the ICE table) because the initial concentrations are quite high.

2007-03-15 01:26:43 · answer #1 · answered by bellerophon 6 · 0 0

Whatever you get for your H+ you take the -log of that and that gives you the pH. That is if you did your ice table correct. Did you multiply 246.5mL times .79 M divided by total mL s that answer is your lactic acid in the ice table, then you setup your ice table ka=products/ reactants

2007-03-14 22:15:23 · answer #2 · answered by Anonymous · 0 0

What I hate about exercises like this one is that a real chemist would always just mix the two chemicals and MEASURE the pH, no need to waste time predicting it.

2007-03-14 21:37:10 · answer #3 · answered by Dorkus 2 · 0 0

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