Question

Make 350 mL of 43 mM sodium phosphate buffer, pH 7.0 without titrating with NaOH or HCI. Show your calculations, which 2 species of sodium phosphate you choose (species are H3PO4, NaH2PO4, Na2HPO4, and Na3PO4; dissociation pKa’s are 2.16, 7.21, and 12.32), and how much of each you weighed out. Check and read your final pH.

Answer 1

i) Following are the dissociation equilibriums of H3PO4

H3PO4(aq) + H2O(l) <-----> H2PO4^-(aq) + H3O⁺(aq)

pKa₁ = 2.16

H2PO4^-(aq) + H2O(l) <-------> HPO4^2-(aq) + H3O⁺(aq)

pKa₂ = 7.21

HPO4^{​​​​​​2-}(aq) + H2O(l) <------> PO4^3-(aq) + H3O⁺(aq)

pKa₃ = 12.32

target pH 7.0 is near to pK₂ , so the required species are H2PO4^- and HPO4^2- and the compounds required are NaH2PO4 and Na2HPO4

ii) Henderson - Hasselbalch which give pH of an buffer solution is

pH = pKa + log([A^-]/[HA])

where,

A^- = conjucate base , HPO4^2-

HA = weak acid , H2PO4^-

pKa of H2PO4^- = 7.21

^{​​​​​​}7.00 = 7.21 + log ([HPO4^-]/[H2PO4^2-])

log([HPO4^2-]/[H2PO4^-]) = -0.21

[HPO4^2-] / [H2PO4^-] = 0.6166

[HPO4^2-] = 0.6166 × [H2PO4^-]

moles of HPO4^2- = 0.6166 × moles × H2PO4^-

molarity of buffer = 43mM = 0.043M

total moles of phosphate species in the buffer = (0.043mol/1000ml) ×350ml = 0.01505mol

moles of H2PO4^- + moles of HPO4^2- = 0.01505 mol

moles of H2PO4^- + (0.6166 × moles of H2PO4^-) = 0.01505mol

1.6166× moles of H2PO4^- = 0.01505mol

moles of H2PO4^- required = 0.0093097

moles of HPO4^2- required = 0.01505mol - 0.0093097mol = 0.0057403mol

moles of NaH2PO4 required = 0.0093097mol

moles of NaHPO4 required = 0.0057403mol

mass = number of moles × molar mass

mass of NaH2PO4 required = 0.0093097mol × 119.98g/mol = 1.1170g

mass of Na2HPO4 required = 0.0057403mol × 141.96g/mol = 0.8149g

Therefore,

mass of NaH2PO4 should be weighed = 1.1170g

mass of Na2HPO4 should be weighed = 0.8149g

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