Question

In: Chemistry

You must make a sodium phosphate buffer (500 ml at 50 mM) at a pH value...

You must make a sodium phosphate buffer (500 ml at 50 mM) at a pH value of 6.8. Assume that the pKa is 7.0. MW oh NaH2PO4 = 138g/mol. MW of NaH2PO4 = 268 g/mol

Show the weak acid base reaction with proper structures. use the Henderson/ Hasselbalch relationship to obtain molarity of each species. Calculate the grams of each salt required.

Expert Solution

I have a little doubt, but I'm almost sure this is the correct answer:

You need a buffer of sodium phosphate, so the combination of sodium phosphate and H3PO4 must be equal to 50 mM:

[H₂PO₄^-] + [HPO₄^2-] = 50 mM or 0.05 M

The acid base reaction:

H₃PO₄ ----------> H₂PO₄^- + H⁺

H₂PO₄^- ----------> HPO₄^2- + H⁺

HPO₄^2- ----------> PO₄^3- + H⁺

and the Henderson Hasselbach reaction is:

pH = pKa + log[H₂PO₄^-] / [H₃PO₄]

With this equation we can obtaint the relation between the salt and acid:

7 = 6.8 + log[HPO₄^-] / [H₂PO₄]

7 - 6.8 = log[HPO₄^-] / [H₂PO₄]

10^0.2 = [HPO₄^-] / [H₂PO₄]

[HPO₄^-] / [H₂PO₄] = 1.5849

[HPO₄^-] = 1.5849 [H₂PO₄]

If [H₂PO₄^-] + [HPO₄] = 50 mM or 0.05 M, then:

1.5849 [H₂PO₄] + [H2PO₄] = 0.05 M

2.5849 [H2PO₄] = 0.05

[H₂PO₄] = 0.0193 M

Then: [HPO₄^-] = 1.5849 * 0.0193 = 0.0306 M

moles of HPO₄^- = 0.0306 * 0.5 = 0.0153 moles

moles of H₂PO₄ = 0.0193 * 0.5 = 0.00965 moles

Finally the mass:

mH₂PO₄^- = 0.0153 * 138 = 2.1114 g

mHPO₄ = 0.00965 * 268 = 2.5862 g

Hope this helps

queen_honey_blossom answered 2 years ago

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