Question

A 1.33 L buffer solution consists of 0.274 M propanoic acid and 0.200 M sodium propanoate....

A 1.33 L buffer solution consists of 0.274 M propanoic acid and 0.200 M sodium propanoate. Calculate the pH of the solution following the addition of 0.065 mol HCl. Assume that any contribution of the HCl to the volume of the solution is negligible. The Ka of propanoic acid is 1.34×10−5

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Answer #1


mol of HCl added = 0.065 mol

C2H5COO- will react with H+ to form C2H5COOH

Before Reaction:
mol of C2H5COO- = 0.2 M *1.33 L
mol of C2H5COO- = 0.266 mol

mol of C2H5COOH = 0.274 M *1.33 L
mol of C2H5COOH = 0.3644 mol

after reaction,
mol of C2H5COO- = mol present initially - mol added
mol of C2H5COO- = (0.266 - 0.065) mol
mol of C2H5COO- = 0.201 mol

mol of C2H5COOH = mol present initially + mol added
mol of C2H5COOH = (0.3644 + 0.065) mol
mol of C2H5COOH = 0.4294 mol


Ka = 1.34*10^-5

pKa = - log (Ka)
= - log(1.34*10^-5)
= 4.873

since volume is both in numerator and denominator, we can use mol instead of concentration
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.873+ log {0.201/0.4294}
= 4.543

Answer: 4.54

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