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Determine the mass (in g) of sodium butanoate (NaC3H7COO) that must be added to 78.9 mL...

Determine the mass (in g) of sodium butanoate (NaC3H7COO) that must be added to 78.9 mL of 0.609 M butanoic acid to yield a pH of 6.43. Report your answer to 3 significant figures.



Assume the volume of the solution does not change and that the 5% approximation is valid.

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Answer #1

pH = 6.43

pKa = 4.824

moles of butanoic acid = 78.9 x 0.609 / 1000

                                     = 0.04805 mol

pH = pKa + log [salt / acid]

6.43 = 4.824 + log [NaC3H7COO / 0.04805]

NaC3H7COO = 1.94

moles of NaC3H7COO = 1.94

mass of NaC3H7COO = 1.94 x 110.1

mass of NaC3H7COO = 213 g

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