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.
4 answers
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.
@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?
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?
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.
I just responded to your first post down below about the pH 6 thing. The absolute value is not the way to go.
1 answer