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.
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
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1 answer