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
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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