Butanoic (butyric) acid, C3H7COOH, is a monoprotic acid with Ka = 1.51×10–5. A 35.00 mL sample of 0. — Acid-Base / Titration Chemistry Question
Problem Context
Butanoic (butyric) acid, C3H7COOH, is a monoprotic acid with Ka = 1.51×10–5. A 35.00 mL sample of 0.500 M butanoic acid is titrated with 0.200 M KOH.
Calculate the [H+] in the original butanoic acid solution.
Model Answer
Let HA = C3H7COOH, and A–= C3H7COO–
Let [H+] and [A-] = x. Plugging in we get,
Solving for x gives, [H+] = 2.73×10–3
Calculate the pH after 10.00 mL of KOH have been added.
Model Answer
Determine the initial number of moles of acid:
Determine the number of moles of NaOH added:
Determine the molarity of HA: 0.0175 – 0.0020 = 0.0155 mol HA remain, in 0.045 L, so the molarity is
The [A-] is changed only through dilution, [A-] =
Plug these values into the equilibrium constant expression and solve for [H+].
and [H+] = 1.16×10-4, so pH = -log(1.16×10-4) = 3.93
Determine the pH at the half-equivalence point of the titration.
Model Answer
At the half equivalence point: [HA] = [A-] and [H+] = Ka = 1.51×10–5. So, pH = -log(1.51×10–5) = 4.82
Find the volume of KOH solution needed to reach the equivalence point for the titration.
Model Answer
The initial moles HA (from part b.) = 0.0175 mol HA, so the equivalence point is reached when we have 0.0175 mol OH– added.
= 0.0875 L.
Calculate the pH at the equivalence point.
Model Answer
Total volume at the equivalence point is 0.0875 L + 0.0350 L = 0.1225 L
At the equivalence point all HA is converted to A-, so:
The A- is a base according to the equation, A– + H2O ⇌ HA + OH–, and
Let [OH–] and [HA] = x. Plugging in we get,
Solving for x = [OH–] = 9.73×10-6.
So, pOH = -log(9.73×10-6) = 5.01 and pH = 14.00 – 5.01 = 8.99