To simulate blood conditions, a phosphate
buffer system with a pH = 7.4 is desired.
What mass of Na2HPO4 must be added to
0.25 L of 0.11 M NaH2PO4(aq) to prepare
such a buffer?
Answer in units of g.
pH = pKa + log (base/acid) The Ka of H2PO41- is 6.2 x 10^-8 pKa of H2PO41- is -log (6.2 x 10^-8) = 7.21 7.40 = 7.21 + log ([HPO42-] / 0.11) 0.19 = log ([HPO42-] - log (0.11) 0.19 + log (0.11) = log [HPO42-] 0.19 - 0.96 = -0.77 = log [HPO42-] [HPO42-] = 10^-0.77 = 0.17 M M = mol / L mol = M x L = 0.17 mol / L x 0.25 L = 0.04 mol Na2HPO4
0.04 mol Na2HPO4 x (142 g / 1 mol ) = 5.68 g Na2HPO4 to be added to 0.25 L of 0.11 M NaH2PO4
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