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How much KH2PO4 should be weighed out to yield a buffer with concentration of 100 mM...

How much KH2PO4 should be weighed out to yield a buffer with concentration of 100 mM phosphate in 100 mL? Report your answer to 2 decimal places.

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Answer #1

Write down the acid dissociation reaction as below.

H2PO4- (aq) --------> H+ (aq) + HPO42- (aq); pKa = 7.20

The pH is not mentioned; hence, I can only figure out the mass of KH2PO4 required to prepare a 100 mM solution.

Molar mass of KH2PO4 = (1*39.0983 + 2*1.008 + 1*30.9738 + 4*15.9994) g/mol = 136.0857 g/mol.

Mole(s) of KH2PO4 in the buffer = (100 mL)*(1 L/1000 mL)*(100 mM)*(1 M/1000 mM)*(1 mol/L/1 M) = 0.01 mole.

Mass of KH2PO4 required = (0.01 mole)*(136.0857 g/1 mole) = 1.360857 g ≈ 1.36 g (ans).

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