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.

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.

You are correct in identifying disodium phosphate (Na2HPO4) as the conjugate base (A-) and monosodium phosphate (NaH2PO4) as the conjugate acid (HA). To determine the pH of the buffer, we need to know the pKa of the conjugate acid-base pair.

The pKa is a measure of the acidity of an acid, and specifically, it is the pH at which half of the acid is dissociated to its conjugate base. In this case, the equilibrium is:

HA ⇌ H+ + A-

To find the pKa for this particular pair of HA and A-, we can refer to tables or databases. In this case, the pKa for monosodium phosphate (NaH2PO4) is 7.21 at 25°C. It is important to mention that the pKa values might vary slightly depending on the temperature.

Once we have the pKa value, we can use the Henderson-Hasselbalch equation to calculate the pH of the buffer:

pH = pKa + log10 ([A-]/[HA])

Given that the final concentration of monosodium phosphate (HA) is 0.3 M and the final concentration of disodium phosphate (A-) is 0.5 M, we can substitute the values into the equation and solve for the pH:

pH = 7.21 + log10 (0.5/0.3)

Using logarithmic rules, we can simplify the equation further:

pH = 7.21 + log10 (1.67)

By evaluating log10 (1.67), which is approximately 0.2227, we can calculate the pH of the buffer:

pH ≈ 7.21 + 0.2227

pH ≈ 7.43

Therefore, the pH of this buffer solution is approximately 7.43.