Question

In the titration of 25.00 mL of 0.100 M acetic acid, what is the pH of the resultant solution after the addition of 25.00 mL of 0.100 M KOH? Assume that the volumes of the solutions are additive. Ka = 1.8x10^-5 for CH₃COOH.

Answer 1

The balanced chemical equation for your neutralization reaction was

CH3COOH(aq)+OH−(aq)→CH3COO−(aq)+H2O(l)

Calculate how many moles of acetic acid and of potassium hydroxide you add together

C=nV⇒n=C⋅V

nCH3COOH=0.100 M⋅25⋅10−3L=0.0025 moles

nNaOH=0.100 M⋅25⋅10−3L=0.0025 moles

Since you add equal numbers of moles of acetic acid and potassium hydroxide, bothcompounds will be completely consumed by the reaction.

In the process, the reaction will produce 0.0025 moles of acetate anions, CH3COO−.

nCH3COOH=0.0025−0.0025=0

nKOH=0.0025−0.0025=0

nCH3COO−=0.0025 moles

The acetate anion can act as a weak base and react with water to produce hydroxide ions and reform acetic acid according to the following equilibrium

CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq)

To determine the concentration of the acetate anion, use the total volume of the solution, which is

Vsol=Vacetic acid+Vpotassium hydroxide

Vsol=25+25=50. mL

This will get you

[CH3COO−]=0.0025 moles50.⋅10−3L=0.0500 M

Use an ICE table to determine the concentration of the hydroxide ions in solution

CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq)
I.......0.0500.............................................0.............................0
C........(-x).................................................(+x)..........................(+x)
E......0.0500-x..........................................x..............................x

By definition, the base dissociation constant, Kb, will be (ka.kb= 1x 10 ^-14) then Kb=5.6X10⁻¹⁰

Kb=[OH−]⋅[CH3COOH]/ [CH3COO−]

= x⋅x/0.0500−x= x²/0.0500−x

Since Kb is so small, you can approximate (0.0500 - x) with 0.0500. This will get you

Kb = x² /0.0500=5.6⋅10−10⇒ x= 5.29⋅10−6

Since x=[OH−], you have

[OH−]=5.29⋅10−6M

The pOH of the solution will be

pOH=−log([OH−])=−log(5.29⋅10−6)=5.28

Therefore, the pH of the solution will be

pHsol=14−pOH=14−5.28=8.72