Answer:
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!
