A metilamina
é um derivado da base fraca do amoníaco. Escreva a equação para a ionização da metilamina na água, encontrar o pH de uma solução de
a
.
a
.
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Uma solução-tampão é preparada a partir de
de
e
de solução aquosa
de
. Sendo assim, determine: Dados:(H=1;
a) o pH desta solução;
b) o pH após adição de 700 mL de água destilada à solução-tampão, justificando com os cálculos. Dados:
<img class=“imgSvg” id = “m1gnc62pjruil632y5” 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Uma solução-tampão é preparada a partir de
a) o pH desta solução;
b) o pH após adição de 700 mL de água destilada à solução-tampão, justificando com os cálculos. Dados:
Passo 1
Beleza, vamos resolver esse exercício de química sobre solução-tampão! A gente tem 6,4 g de NH₄NO₃ e 0,10 L de uma solução aquosa de NH₄OH com concentração de 0,080 mol/L. A gente precisa descobrir o pH dessa solução. Para isso, vamos precisar calcular os moles das substâncias envolvidas e usar a constante de base (Kb).
Passo 2
Primeiro, vamos calcular quantos moles de NH₄NO₃ temos. A fórmula do NH₄NO₃ é formada por NH₄⁺ e NO₃⁻. A massa molar do NH₄NO₃ é:
- N: 14 g/mol
- H: 1 g/mol
- O: 16 g/mol
Então:
Massa molar do NH₄NO₃ = (2×14) + (4×1) + (3×16) = 28 + 4 + 48 = 80 g/mol
Massa molar do NH₄NO₃ = (2×14) + (4×1) + (3×16) = 28 + 4 + 48 = 80 g/mol
Agora, calculamos os moles:
Moles de NH₄NO₃ = massa / massa molar = 6,4 g / 80 g/mol = 0,08 mol
Moles de NH₄NO₃ = massa / massa molar = 6,4 g / 80 g/mol = 0,08 mol
Passo 3
Agora, vamos calcular os moles de NH₄OH. A concentração é 0,080 mol/L e o volume é 0,10 L.
Moles de NH₄OH = concentração × volume = 0,080 mol/L × 0,10 L = 0,008 mol
Passo 4
Na solução-tampão, temos NH₄⁺ de NH₄NO₃ e NH₄OH funcionando como a base. A reação de equilíbrio é:
Para encontrar o pH, vamos usar a expressão da constante de base (Kb) para NH���OH:
Como temos uma solução tampão, podemos usar a fórmula de Henderson-Hasselbalch para ácidos e bases fracas:
Primeiro, precisamos encontrar o pKa relacionado ao Kb dado:
Agora, vamos identificar as concentrações de ácido e base na solução:
- [Ácido] = [NH₄⁺] = 0,08 mol / 0,10 L = 0,8 M
- [Base] = [NH₃] ≈ [NH₄OH] = 0,008 mol / 0,10 L = 0,08 M
Passo 5
Aplicando a fórmula de Henderson-Hasselbalch:
Resposta
O pH da solução-tampão é 8,25.