Answer:
pH → 7.47
Explanation:
Caffeine is a sort of amine, which is a weak base. Then, this pH should be higher than 7.
Caffeine + H₂O ⇄ Caffeine⁺ + OH⁻ Kb
1 mol of caffeine in water can give hydroxides and protonated caffeine.
We convert the concentration from mg/L to M
415 mg = 0.415 g
0.415 g / 194.19 g/mol = 2.14×10⁻³ mol
[Caffeine] = 2.14×10⁻³ M
Let's calculate pH. As we don't have Kb, we can obtain it from pKb.
- log Kb = pKb → 10^-pKb = Kb
10⁻¹⁰'⁴ = 3.98×10⁻¹¹
We go to equilibrium:
Caffeine + H₂O ⇄ Caffeine⁺ + OH⁻ Kb
Initially we have 2.14×10⁻³ moles of caffeine, so, after the equilibrium we may have (2.14×10⁻³ - x)
X will be the amount of protonated caffeine and OH⁻
Caffeine + H₂O ⇄ Caffeine⁺ + OH⁻ Kb
(2.14×10⁻³ - x) x x
We make the expression for Kb:
3.98×10⁻¹¹ = x² / (2.14×10⁻³ - x)
We can missed the -x in denominator, because Kb it's a very small value.
So: 3.98×10⁻¹¹ = x² / 2.14×10⁻³
√(3.98×10⁻¹¹ . 2.14×10⁻³) = x → 2.92×10⁻⁷
That's the [OH⁻]. - log [OH⁻] = pOH
- log 2.92×10⁻⁷ = 6.53 → pOH
14 - pOH = pH → 14 - 6.53 = 7.47
