How to calculate magnitude of k
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Answer:
The [SO₃²⁻]
Explanation:
From the first dissociation of sulfurous acid we have:
H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)
At equilibrium: 0.50M - x x x
The equilibrium constant (Ka₁) is:
With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:
Similarly, from the second dissociation of sulfurous acid we have:
HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)
At equilibrium: 7.94x10⁻²M - x x x
The equilibrium constant (Ka₂) is:
Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:
Therefore, the final concentrations are:
[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M
[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M
[SO₃²⁻] = 7.07x10⁻⁵M
[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M
So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.
I hope it helps you!
Answer:
El potencial celular estándar, is +0.46 V
Explanation:
Las reacciones de media célula son;
Media reacción del ánodo Cu²⁺ + 2e⁻ ↔ Cu, E ° = 0.34 V
Media reacción catódica 2Ag + 2e⁻ ⁻ 2Ag, E ° = 0.80 V
Sin embargo tenemos para hierro Fe²⁺ + 2e⁻ ↔ Fe, E ° -0.44 V
y Fe³⁺ + e⁻ ↔ Fe²⁺, E ° = 0.77 V
que es más alta que la del cobre presente, por lo tanto, el cobre se oxidará en el ánodo
Por lo tanto, en el ánodo, tendremos
Cu → Cu²⁺ + 2e⁻ (E ° = -0.34 V)
En el cátodo
2Ag + 2e⁻ → 2Ag (E ° = 0.80 V)
El potencial celular estándar, = +0.46 V