A 1.24-g sample of benzoic acid was dissolved in water to give 50.0 mL of solution. This solution was titrated with 0.180 M NaOH. What was the pH of the solution when the equivalence point was reached?

pH = 8.59

Explanation:

1,24g of benzoic acid are:

1,24g × (1mol / 122,12g) = 0,0102 moles

Benzoic acid in water is in equilibrium with benzoate ion, thus:

Benzoic acid ⇄ Benzoate⁻ + H⁺. K =  [Benzoate⁻] [H⁺] / [Benzoic acid] = 6,3x10⁻⁵

Also, benzoic acid reacts with NaOH producing benzoate ion and water:

Benzoic acid + NaOH → Benzoate⁻ + H₂O + Na⁺

The volume spent during titration is:

0,0102mol × (1L / 0,180mol) = 0,0567L. That means total volume is 0,0567L + 0,0500L = 0,1067L.

The equivalence point is reached when all benzoic acid reacts with NaOH and there is just benzoate. That is molarity of benzoate is 0,0102mol / 0,1067L = 0,0956M

The equilibrium of benzoate ion is:

Benzoate⁻ + H₂O ⇄ Benzoic acid + OH⁻ kb = kw/ka = 1x10⁻¹⁴/6,3x10⁻⁵ = 1,59x10⁻¹⁰ = [Benzoic acid] [OH⁻] / [Benzoate⁻]

The equilibrium concentrations are:

[Benzoate⁻] = 0,0956M - x

[Benzoic acid] = x

[OH⁻] = x

Replacing:

1,59x10⁻¹⁰ = x² / 0,0956M - x

1.52x10⁻¹¹ - 1.59x10⁻¹⁰x - x² = 0

Solving for x:

X = -3.8988×10⁻⁶ Wrong answer, there is no negative concentrations-

X = 3.89864×10⁻⁶

Thus:

[OH⁻] = 3.89864×10⁻⁶

pH = 14 - pOH and pOH = -log [OH⁻]

pOH = -log 3.89864×10⁻⁶ = 5,41

pH = 14 - 5,41 = 8,59

pH = 8.59

I hope it helps!

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