You decided to prepare a phosphate buffer from solid sodium dihydrogen phosphate (nah2po4) and disodium hydrogen phosphate (na2hpo4)and you need 1l of the buffer at ph 7.00 with a total phosphate concentration(nah2po4+ na2hpo4) of 0.100 m. hint: phosphoric acid (h3po4), a triprotic acid, has 3 pka values: 2.14, 6.86, and12.4. determine the weight in grams of sodium dihydrogen phosphate (nah2po4; 138g/mol) and disodium hydrogen phosphate (na2hpo4; 142g/mol) needed to prepare this buffer.

For disodium hydrogen phosphate:

5.32g Na2HPO4

For sodium dihydrogen phosphate:

7.65g Na2HPO4

Explanation:

First, you have to put all the data from the problem that you going to use:

-NaH2PO4 (weak acid)

-Na2HPO4 (a weak base)

-Volume = 1L

-Buffer pH = 7.00

-Concentration of [NaH2PO4 + Na2HPO4] = 0.100 M

What we need to find the pKa of the weak acid, in this case NaH2PO4, for that you need to find the Ka (acid constant) of NaH2PO4, and for this we use the pKa of the phosphoric acid as follow:

H3PO4 = H2PO4 + H+    pKa1 = 2.14

H2PO4 = HPO4 + H+       pKa2 = 6.86

HPO4 = PO4 + H+      pKa3 = 12.4

So, for the preparation of buffer, you need to use the pKa that is near to the value of the pH that you want, so the choice will be:

pKa2= 6.86

Now we going to use the Henderson Hasselbalch equation for the pH of a buffer solution:

pH = pKa2 + log [(NaH2PO4)/(Na2HPO4)]

The solution of the problem is attached to this answer.

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b