Question

If a liter of a buffer is prepared with a final concentration of .3M monosodium phosphate and .5M disodium phosphate, what is the pH of this buffer? (Hint: Which is conjugate acid (HA) and conjugate base (A-)? What's the pKa for this particular pair of HA and A-?)

So I think the conjugate base HA is disodium phosphate and the conjugate acid A- is monosodium phosphate. Otherwise I'm not sure where to start and I'm not sure why or how I determine that at this point.

Answer 1

You want to use the Henderson-Hasselbalch equation. Substitute into
pH = pKa + log[(base)/(acid)].
The base is Na2HPO4. The acid is NaH2PO4.

Answer 2

Okay, so based on the chart in my book I was able to determine the following.

pH=(unknown)
pKa=7.2 of Phosphoric Acid (2)
A-=.5M of Na2HPO4
HA=.3M H2PO4-
pH= 7.2 + log(.5/.3)
pH=7.2+log(1.6667)
pH=7.2+0.5108

So, the pH of this solution is:
pH=7.7108

Correct?

Answer 3

pH=(unknown)
pKa=7.2 of Phosphoric Acid (2)
A-=.5M of Na2HPO4
HA=.3M H2PO4-
pH= 7.2 + log(.5/.3)
pH=7.2+log(1.6667)
<b>You are ok to here</b>
pH=7.2+0.5108
<b>I find 0.22 for the log of 1.6667</b>

Answer 4

Upon recalculating this you are definitely correct...log(1.6667)=0.22
so 7.2+.22=7.42
and the pH is then 7.42
Thanks again for your help.