Thank you!!!! Phosphoric acid is a triprotic acid with the following pK, values:

Question

Thank you!!!! Phosphoric acid is a triprotic acid with the following pK, values: pK., 2.148 pK 7.198 pK, 12.375 You wish to prepare 1.000 L of a 0.0300 M phosphate buffer at pH 7.740. 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). HsPO4 and NaH2PO4 H3PO4 and NazHPO4 H3PO4 and Na3PO4 NaH2PO4 and NayPO4 Na2HPO4 and NajPO

Explanation / Answer

ans)

Firstly you need to do the calculations via Handerson Hasselbach equation:

There are three pKa's of phosphoric acid. If required pH is 7.74, 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, 7.74 = 7.198 + log([A]/[HA])
log([A]/[HA]) = 0.542
[A]/[HA] = 3.48

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=0.03
A/HA= 3.48

HA=0.006 M
A=0.024 M

This means, you need to put 0.006 moles of NaH2PO4 and 0.024 moles of Na2HPO4 salts into 1 liters of solution.

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

Thus, 0.006 x 120 = 0.72 g of NaH2PO4 and 0.024x 142 = 3.408 g of Na2HPO4 should be added

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