Asked by Mike

A biochemist prepares a lactic acid-lactate buffer by mixing 225 mL of 0.85M lactic acid with a Ka=1.38*10^-4 with 435mL of 0.68M sodium lactate. What is the buffer pH?

I know I have to use the equilibrium constant expression, but I'm pretty lost from there. Can someone please explain this for me? Thanks!
Answered by DrBob222
No, this is a buffer problem. Use the Henderson-Hasselbalch equation.
225 mL x 0.85M lactic acid(HL) = about 199.75 millimols acid.

435 mL x 0.68M = about 295.8 mmols NaL which is the base.

pH = pKa + log (base)/(acid)

Answered by Mike
Ohhhh, that makes sense! Thanks!
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