Many of the biochemistry experiments and cell cultures need to be pH buffered in

Question

Many of the biochemistry experiments and cell cultures need to be pH buffered in a narrow range. Many different buffers are used, but phosphate buffers are a classic choice. Suppose you need to make a 1 liter phosphate buffer solution with a pH of 7.10. You add 15.0 grams of sodium dihydrogenphosphate (NaH2PO4) to a 1.00 liter flask. How many grams of sodium hydrogen phosphate (Na2HPO4) do you need to add to obtain the desired pH buffer solution? The pKa values for the three ionization states of phosphoric acid are 2.16, 7.21 and 12.32. The exacting nature of these types of solutions requires that you be within 0.25 grams to receive credit for this question.

Explanation / Answer

Solution :-

15.0 g NaH2PO4

Lets first calculate the moles of the NaH2PO4

Moles = mass / molar mass

Moles of NaH2PO4 = 15.0 g / 119.98 g per mol = 0.125 mol

H2PO4^- will acts as acid and HPO4^2- will acts as base in the buffer

So we have to find the mass of Na2HPO4 needed to form with pH 7.10

So lets use the Henderson equation

pH= pka + log ([base]/[acid])

need to use the second dissociation constant so use the pka2 that is 7.21

7.10 = 7.21 + log ([HPO4^2-] /[0.125 M] )

7.10 – 7.21 = log ([HPO4^2-] /[0.125 M] )

-0.11 = log ([HPO4^2-] /[0.125 M] )

Antilog [-0.11] = [HPO4^2-] /[0.125 M]

0.7762 = [HPO4^2-] /[0.125 M]

[HPO4^2-] = 0.7762 * 0.125 M

[HPO4^2-] = 0.097 M

Since volume is 1.00 L

Therefore moles of HPO4^2- needed = 0.097 mol

So the moles of Na2HPO4 needed = 0.097 mol

Now lets calculate its mass

Mass = moles * molar mass

          = 0.097 mol *141.96 g per mol

         = 13.8 g Na2HPO4

So the mass of the Na2HPO4 that is necessary to add to the solution is 13.8 g   

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