A chemist starts with 1.00 L of 0.250 M propanoic acid, CH₃CH₂COOH ( at 25 °C)
CHEMISTRY · Multi-part · A Level · Chemistry · Acids and bases · Acids and bases
Problem
A chemist starts with 1.00 L of 0.250 M propanoic acid, CH₃CH₂COOH (pKa=4.87 at 25 °C). Solid NaOH pellets (assume 100 % purity) will be added to convert some of the acid into its conjugate base so that, after complete dissolution of NaOH and negligible change in total volume, the solution attains a pH of 5.20.
(a) Determine the required value of [CH3CH2COOH][CH3CH2COO−] in the final solution.
(b) Calculate the initial number of moles of propanoic acid present before any NaOH is added.
(c) Show how many moles of propanoic acid must react with NaOH to produce the conjugate base in the proportion found in part (a).
(d) Compute the mass of solid NaOH (in grams) that must be weighed out to achieve the target pH, assuming the final volume remains effectively 1.00 L.
(e) State one experimental reason, aside from calculation error, why the actual pH obtained in the laboratory might differ slightly from the predicted value.