Question

While carring out a research project you find that you need a 0.15 M solution of sodium phosphate buffer, at pH of 8.4. Unfortunately your technician has left already and you must make the solution yourself. On the shelf you find bottles labeled: NaH2PO4.2H2O Wt. 156.01, and Na2HPO4 Wt. 141.96. What weights of A and B in grams would you dissolve to make 1 liter of the required buffer?( The pKs for phosphoric acid are 2.0, 7.2 and 12.0).

Answer 1

Use the Hendderson Hasselbach equation:

pH = pKa + log [S]/[A]

Now if you need to make the buffer, this means that you need to sum both solutions, let's call the Buffer as "B" so:

[B] = [NaH₂PO₄] + [Na₂HPO₄]

As you need to make a buffer of 8.4, you need to use the value of pKa closer to that one, in this case, the second one according to the following reaction of phosphoric acid:

H₂PO₄^- -----------> HPO₄^2- + H⁺   pKa2 = 7.4

So the HH equation would be: pH = pKa + log [HPO₄^2-] / [H₂PO₄^-]

8.4 = 7.2 + log  [HPO₄^2-] / [H₂PO₄^-]

8.4 - 7.2 = log  [HPO₄^2-] / [H₂PO₄^-]

10^1.2 =  [HPO₄^2-] / [H₂PO₄^-]

[HPO₄^2-] / [H₂PO₄^-] = 15.85

[HPO₄^2-] = 15.85[H₂PO₄^-]

0.15 = [NaH₂PO₄] + [Na₂HPO₄]

0.15 = 15.85[H₂PO₄^-] + [H₂PO₄^-]

0.15 / 16.85 = [H₂PO₄^-]

[H₂PO₄^-] = 0.0089

[HPO₄^2-] = 15.85 * 0.0089 = 0.1411 M

mass HPO₄^2- = 0.1411 mol/L * 1 L * 141.96 g/mol = 20.0306 g

mass H₂PO₄^- = 0.0089 * 1 * 156.01 = 1.3885 g

Hope this helps