As a technician in a large pharmaceutical research firm, you need to produce 400

Question

As a technician in a large pharmaceutical research firm, you need to produce 400. mL of a potassium dihydrogen phosphate buffer solution ofpH = 6.84. The pKa of H2PO4 is 7.21.

You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O.

How much 1.00 M KH2PO4 will you need to make this solution? (Assume additive volumes.)

Express your answer to three significant digits with the appropriate units.

Volume of KH2PO4 needed =

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70.6 is wrong answer !!!!!! if you get this answer do not waste my question

Explanation / Answer

Use the buffer equation:

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

Here:

log([HPO42-]/[H2PO4-]) = pH- pKa = 6.84 – 7.21 = - 0.37

[HPO42-]/[H2PO4-] = 10 -0.37= 0.4266

      The two phosphate solutions have the same concentration, thus the ratio 0.4266 :1

   is obtained by mixing the volumes in this ratio.

            VKH2PO4 = 400 mL x 1/ (1+0.4266) = 280 mL

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