Posted by Shaun on Wednesday, July 6, 2011 at 11:13am.
You want to use the Henderson-Hasselbalch equation. Substitute into
pH = pKa + log[(base)/(acid)].
The base is Na2HPO4. The acid is NaH2PO4.
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?
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)
You are ok to here
pH=7.2+0.5108
I find 0.22 for the log of 1.6667
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.
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