What volumes of 0.5 M Sodium phosphate monobasic (NaH2PO4), 0.5 M Sodium phosphate dibasic ( Na2HPO4) and water would you have to mix to make 500mL of a 0.250M Phosphate buffer at pH 7.6?
Answer
Volume of 0.5M NaH2PO4 required = 72.36ml
Volume of 0.5M Na2HPO4 required = 177.64ml
Volume of water required = 250ml
Explanation
Henderson - Hasselbalch equation is
pH = pKa + log([A-]/[HA])
A- = HPO42-
HA = H2PO4-
pKa of H2PO4- = 7.21
7.60 = 7.21 + log([HPO42-]/[H2PO4-])
log([HPO42-]/[H2PO4-]) = 0.39
[HPO42-]/[H2PO4-】= 2.455
[HPO42-] = 2.455 × [H2PO4-]
moles of HPO42- = 2.455 × moles of H2PO4-
moles of buffer = (0.250mol/1000ml) × 500ml = 0.125mol
moles of HPO42- + moles of H2PO4- = 0.125mol
2.455 ×moles of H2PO4- + moles of H2PO4- = 0.125mol
3.455×moles of H2PO4- = 0.125mol
moles of H2PO4- = 0.03618mol
moles of HPO42- = 0.125mol - 0.03618mol = 0.08882mol
Volume of 0.5M NaH2PO4 required = (1000ml/0.5mol) × 0.03618mol = 72.36ml
Volume of 0.5M Na2HPO4 required = (1000ml/0.5mol) × 0.08882mol = 177.64ml
Volume of water required= 500ml - ( 72.36ml + 177.64ml) = 250ml
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