You are instructed to create 800. mL of a 0.66 M phosphate buffer with a pH of 6

Question

You are instructed to create 800. mL of a 0.66 M phosphate buffer with a pH of 6.9. You have phosphoric acid and the sodium salts NaH2PO4 Na2HPO4, and Na3PO4 available. (Enter all numerical answers to three significant figures.) H3PO4(s)H2oH30+(aq)H2PO4-(aq) H2PO4-(aq) + H2O(/)-HaO+(aq) + HP042-(aq) HP042-(aq) + H2O(I) ; HaO+(aq) + PO43-(aq) K., = 6.9 x 10-3 Ka2 = 6.2 x 10-8 ??,-4.8 x 10-13 Which of the available chemicals will you use for the acid component of your buffer? H2 PO NaH2 PO4 Na2HPO4 Na PO 3P4 Which of the available chemicals will you use for the base component of your buffer? H2 PO NaH2PO Na2 HPO4 Na PO What is the molarity needed for the acid component of the buffer? What is the molarity needed for the base component of the buffer? How many moles of acid are needed for the buffer? mol How many moles of base are needed for the buffer?

Explanation / Answer

According to Henderson-Hasselbulch equation

pH = pKa2 + Log(nNa2HPO4/nNaH2PO4)

i.e. 6.9 = -Log(6.2*10-8) + Log(nNa2HPO4/nNaH2PO4)

i.e. 6.9 = 7.2 + Log(nNa2HPO4/nNaH2PO4)

i.e. Log(nNaH2PO4/nNa2HPO4) = 7.2 - 6.9 = 0.3

i.e. nNaH2PO4/nNa2HPO4 = 100.3 = 2 ............... Equation 1

Here, nNaH2PO4 + nNa2HPO4 = 800 mL * 0.66 mmol/mL = 528 mmol

i.e. nNaH2PO4 = 800 - nNa2HPO4 ............... Equation 2

From equations 1 and 2

(800 - nNa2HPO4)/nNa2HPO4 = 2

nNa2HPO4 = 800/3 = 266.67 mmol

Therefore, the no. of moles of the base that are needed for the buffer = 0.267 mol

The mass of the base needed for the buffer = 0.267 mol * 142 g/mol = 37.914 g

The molarity needed for the base component of the buffer = 266.67 mmol/800 mL = 0.334 M

From equation 1, nNaH2PO4 = 2*266.67 = 533.34 mmol

Therefore, the no. of moles of the acid that are needed for the buffer = 0.533 mol

The mass of acid needed for the buffer = 0.533 mol * 120 g/mol = 63.96 g

The molarity needed for the acid component of the buffer = 533.34 mmol/800 mL = 0.666 M

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