You wish to prepare 1.000 L of a 0.0400 M phosphate buffer at pH 7.660. To do th

Question

You wish to prepare 1.000 L of a 0.0400 M phosphate buffer at pH 7.660. To do this, you choose to mix the two salt forms involved in the second ionization, NaH2PO4 and Na2HPO4, in a 1.000 L volumetric flask and add water to the mark. What mass of each salt will you add to the mixture?

Phosphoric acid is a triprotic acid with the following pK, values pK 2.148 pK.2-7.198 pK3 2375 You wish to prepare 1.000 L of a 0.0400 M phosphate buffer at pH 7.660. To do this, you choose to mix the two salt forms involved in the second ionization, NaH2PO4 and Na2HPO4, in a 1.000 L volumetric flask and add water to the mark. What mass of each salt will you add to the mixture? Number Number mass NaH,PO, mass Na,HPO,- What other combination of phosphoric acid and/or its salts could be mixed to prepare this buffer? (Check all that apply). HPO4 and NaH2PO4 HyPO4 and Na2HPO4 HPO4 and NaPO4 NaH2PO4 and Na3PO4 Na2HPO4 and Na3PO4

Explanation / Answer

Firstly you need to do the calculations via Handerson Hasselbach equation:
There are three pKa's of phosphoric acid. If required pH is 7.66, then, pKa27.198 will be used. This means, sodium dihydrogen phosphate and sodium monohydrogen phosphate (diprotic (H2PO4-) and monoprotic (HPO4--) potassium salts) will be used.

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

Now, for A- you put Na2HPO4 concentration, and for HA you put NaH2PO4 concentration:

so, 7.66 = 7.198 + log([A]/[HA])
log([A]/[HA]) = 0.462
[A]/[HA] = 2.90

So, now you know the fold difference between these salts, and the Molarity of the solution is given to you, and you can calculate the required amount:

HA+A=0.04
A/HA= 2.90

HA=0.0103 M
A=0.0297 M

This means, you need to put 0.0103 moles of NaH2PO4 and 0.0297 moles of Na2HPO4 salts into 1 liters of solution.

for molecular weights: NaH2PO4= 23+97=120 amu; Na2HPO4= (23x2)+96=142 amu

Thus, 0.0103 x 120 = 1.236 g of NaH2PO4 and 0.0297 x 142 = 4.217 g of Na2HPO4 should be added

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