You are asked to prepare a KH2PO4 – Na2HPO4 solution that has the same pH as hum

Question

You are asked to prepare a KH2PO4 – Na2HPO4 solution that has the same pH as human blood, 7.40.

a) What should be the ratio of concentrations of [HPO42-] / [H2PO4-] in this buffer solution?

b) Suppose you have to prepare 1.00 L of this solution such that it is isotonic with blood (i.e. same osmolarity as blood). What masses of KH2PO4 and of Na2HPO4•12H2O would you use? Assume all of these salts are fully soluble in water. Blood has an osmolarity that is equivalent to 9.2 g NaCl / L. (remember, you are not using NaCl, but merely using this information as a basis of comparison) (HINT: osmolarity is a colligative term which only depend on the total number of solutes formed in solution).

Explanation / Answer

pH = pka + log [HPO42-]/[H2PO4]

7.4= 7.198 + log [HPO42-]/[H2PO4-]

[HPO42-]/[H2PO4-]= 1.5922  

B) [HPO42-] = 1.5922 [H2PO4-].....................(1)

NaCl moles = mass/molar mass = 9.2/58.44 = 0.157426

NaCl gives two ions Na+ and Cl- , hence i = 2 ( i = vantoff factor)

Osmolairty= i x moles of NaCl / sol vol in L = 2 x0.157426 = 0.31485 M

now our buffer iostonic with NaCl solution whch means conc is same

hence [HPO42-] + [H2PO4-] = 0.31485 ....(2)

by (1) ( 2) we get [H2PO4-] = 0.12146     , [HPO42-] = 0.1934

KH2PO4 moles = Mx V = 0.12146 x 1 = 0.12146

Mass = moles x molar mass = 0.12146 x 136 = 16.52 g

Mass of Na2HPO4 .12H2O = moles x molar mas s= 0.1934 x 358.23 =69.3 g

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