What volumes of 0.1M Sodium phosphate monobasic (NaH2PO4), 0.1M sodium phosphate dibasic (Na2HPO4), and water would...
What volumes of 0.1M Sodium phosphate monobasic (NaH2PO4), 0.1M sodium phosphate dibasic (Na2HPO4), and water would you have to mix to make 100mL of a .05M Phosphate buffer at pH 7.4? pKa = 7.21
Homework Answers
Answer
Volume of 0.1M NaH2PO4 required = 19.6ml
Volume of 0.1M Na2HPO4 required = 30.4ml
Volume of water required = 50ml
Explanation
Henderson-Hasselbalch equation is as follows
pH = pKa + log([A-]/[HA])
A- = conjucate base , HPO42-
HA = weak acid , H2PO4-
pKa = 7.21
substituting the values
7.40 = 7.21 + log([HPO42-]/[H2PO4-])
log([HPO42-]/[H2PO4-]) = 0.19
[HPO42-]/[H2PO4-] = 1.549
[HPO42-] = 1.549× [H2PO4-]
[H2PO4-] + (1.549× [H2PO4-]) = 0.05M
2.549[H2PO4-] = 0.05M
[H2PO4-] = 0.05M/2.549
[H2PO4-] = 0.0196M
[ HPO42-] = 0.050M - 0.0196M = 0.0304M
moles of HPO42- required = (0.0304mol/1000ml)×100ml = 0.00304mol
moles of H2PO4- required = (0.0196mol/1000ml)×100ml = 0.00196mol
Volume of 0.1M NaH2PO4 required = (1000ml/0.1mol)×0.00196mol = 19.6ml
Volume of 0.1M Na2HPO4 required = (1000ml/0.10mol) × 0.00304mol = 30.4ml
Volume of water required = 100ml - (19.6ml + 30.4ml) = 50ml
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