You are instructed to create 900. mL of a 0.47 M phosphate buffer with a pH of 6

Question

You are instructed to create 900. mL of a 0.47 M phosphate buffer with a pH of 6.7. You have phosphoric acid and the sodium salts NaH2PO4, Na2HPO4, and Na3PO4available. (Enter all numerical answers to three significant figures.)

Ka1 = 6.9103

Ka2 = 6.2108

Ka3 = 4.81013

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?

How many moles of base are needed for the buffer?

How many grams of acid are needed for the buffer?

How many grams of base are needed for the buffer?

H3PO4(s) + H2O(l) H3O+(aq) + H2PO4(aq)    

Ka1 = 6.9103

H2PO4(aq) + H2O(l) H3O+(aq) + HPO42(aq)    

Ka2 = 6.2108

HPO42(aq) + H2O(l) H3O+(aq) + PO43(aq)

Ka3 = 4.81013

Explanation / Answer

Using Hendersen-Hasselbalck equation,

pH = pKa + log([base]/[acid])

with pH = 6.7, we would need NaH2PO4 and Na2HPO4 as acid and base components

6.7 = 7.2 + log([HPO4^2-]/[H2PO4-])

[HPO4^2-] = 0.32[H2PO4-]

[H2PO4-] + [HPO4^2-] = 0.47 M x 0.9 L = 0.423 mols

[H2PO4-] + 0.32[H2PO4-] = 0.423

Thus,

moles of acid component needed [H2PO4-] = 0.32 mols

moles of base component needed [HPO4^2-] = 0.103 mols

molarity of acid component needed [H2PO4-] = 0.32 mol/0.9 L = 0.35 M

molarity of base component needed [HPO4^2-] = 0.103 mol/0.9 L = 0.11 M

grams of acid component needed = 0.32 mol x 119.98 g/mol = 38.39 g

grams of base component needed = 0.103 mol x 141.96 g/mol = 14.62 g

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