Question

How many moles of conjugate base (Na2HPO4) are present in 40mL of .010M buffer at pH 7.5 vs at pH 6?

Buffer was made up of Na2HPO4 and NaH2PO4.
pH=pKa +log[A/HA]; pKa=7.2.... When I did the calculations I got the same amount as if it were in a 0.1M buffer at the same pH's so I think I'm doing something wrong.

Answer 1

I showed you how to do this for pH 7.5. You do it the same way for pH 6 and the answers are not the same. In instead of us working this again, why not show your work and let us find the error.

Answer 2

@DrBob222 oh, so instead of solving for change in pH we stop after a+b=4mmols. so it'll be different when it goes to .01M.

But 7.5=7.2+logb/a; b=2a a+b=.004mols; 3a=.004 a=.00376 b=.000237. So that would be my final answer for pH 7.5 and I do the same for pH 6?

Answer 3

And, the whole pH 6 thing still doesn't make sense. I just took the absolute value and got a change in pH of .0095.

Answer 4

I just responded to your first post down below about the pH 6 thing. The absolute value is not the way to go.