A 25.0 mL sample of 0.100 M propanoic acid (HC3H5O2, Ka = 1.3 ✕ 10-5) is titrated with 0.100 M KOH solution. calculate the ph when 25 mL of KOH is added
To solve this problem, we need to determine the concentration of propanoic acid (HC3H5O2) and the concentration of the conjugate base (C3H5O2-) at the equivalence point.
1. Determine the number of moles of propanoic acid (HC3H5O2):
moles HC3H5O2 = volume (L) × concentration (M)
moles HC3H5O2 = 0.025 L × 0.100 M = 0.0025 mol
2. Determine the number of moles of KOH that react with propanoic acid:
In a 1:1 stoichiometric ratio, 1 mol of propanoic acid reacts with 1 mol of KOH.
moles KOH = 0.0025 mol
3. Determine the volume of KOH solution required to react with propanoic acid:
moles KOH = volume (L) × concentration (M)
volume KOH = moles KOH / concentration KOH
Given:
moles KOH = 0.0025 mol
concentration KOH = 0.100 M
volume KOH = 0.0025 mol / 0.100 M = 0.025 L = 25 mL
4. At the equivalence point, all the propanoic acid has reacted with KOH to form the conjugate base (C3H5O2-). Therefore, the concentration of the conjugate base is equal to the initial concentration of propanoic acid:
concentration C3H5O2- = 0.100 M
5. To determine the pH at the equivalence point, we need to calculate the concentration of hydroxide ions (OH-) using the concentration of the conjugate base:
[OH-] = Kw / concentration C3H5O2-
Kw (at 298 K) = 1.0 × 10^-14
[OH-] = (1.0 × 10^-14) / 0.100 M = 1.0 × 10^-13 M
6. Calculate the pOH at the equivalence point:
pOH = -log([OH-])
pOH = -log(1.0 × 10^-13) = 13
7. Calculate the pH at the equivalence point:
Since pH + pOH = 14, pH = 14 - pOH
pH = 14 - 13 = 1
Therefore, the pH when 25 mL of KOH is added is 1.