A buffer solution is prepared by mixing 66.3 mL of 0.0409 M sodium dihydrogen ph
ID: 1002931 • Letter: A
Question
A buffer solution is prepared by mixing 66.3 mL of 0.0409 M sodium dihydrogen phosphate with 90.1 mL of 0.855 M sodium hydrogen phosphate.
A table of pKa values can be found here. http://chemresources.chemistry.dal.ca/firstyear/acid_base.html
1. Calculate the pH (to two decimal places) of this solution.
Assume the 5% approximation is valid and that the volumes are additive.
2. Calculate the pH (to two decimal places) of the buffer solution after the addition of 0.361 g of sodium hydrogen phosphate (Na2HPO4) to the buffer solution above.
Assume 5% approximation is valid and that the volume of solution does not change.
Explanation / Answer
Q.1: Moles of sodium dihydrogen phosphate(NaH2PO4) in the buffer solution
= MxV(L) = 0.0409 mol/L x 66.3 mL x (1L / 1000 mL) = 2.711x10-3 mol
Moles of sodium hydrogen phosphate(Na2HPO4) in the buffer solution
= MxV(L) = 0.855 mol/L x 90.1 mL x (1L / 1000 mL) = 7.704x10-2 mol
Total volume of the buffer solution, Vt = 66.3 mL + 90.1 mL = 156.4 mL = 0.1564 L
The buffer action is due to the salt sodium hydrogen phosphate(Na2HPO4) and acid sodium dihydrogen phosphate(NaH2PO4).
The pH of the buffer solution can be calculated from Hendersen equation as
pH = pKa + log[salt]/[Acid] = pKa + log[Na2HPO4] / [NaH2PO4]
=> pH = 7.20 + log(moles of Na2HPO4 / moles of NaH2PO4) [ Since Vt remains same for both]
=> pH = 7.20 + log(7.704x10-2 mol / 2.711x10-3 mol) = 8.65 (answer)
Q.2: Given the extra mass of sodium hydrogen phosphate (Na2HPO4) added = 0.361 g
molar mass of sodium hydrogen phosphate (Na2HPO4) = 141.96 g/mol
Hence moles of sodium hydrogen phosphate (Na2HPO4) added = 0.361 g / 141.96 g/mol = 2.543x10-3 mol
Now pH of the buffer solution can be calculated from Hendersen equation as
pH = pKa + log[Na2HPO4] / [NaH2PO4]
=> pH = 7.20 + log(moles of Na2HPO4 / moles of NaH2PO4) [ Since Vt remains same for both]
=> pH = 7.20 + log[(7.704x10-2 mol +2.543x10-3 mol ) / 2.711x10-3 mol] = 8.67 (answer)
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