If you start with 0.30 m Mn₂ , at 12.5 pH, free Mn₂ concentration be equal to 4.6 x 10⁻¹¹ m
Initial molarity of Mn₂ = 0.30 M
Final molarity of Mn₂ = 4.6 x 10⁻¹¹
pH = ?
Ksp [Mn(OH)₂] = 4.6 x 10⁻¹⁴ (standard value)
Write the ionic equation
Mn(OH)₂ → Mn⁺² + 2OH⁻
[Mn⁺²] = 4.6 x 10⁻¹¹
We will calculate the concentration of OH⁻ by using Ksp expression
Ksp = [Mn⁺²][OH-]²
[Mn⁺²][OH⁻]² = 4.6 x 10⁻¹⁴
[OH⁻]² = 4.6 x 10⁻¹⁴ / 4.6 x 10⁻¹¹
[OH⁻]² = 10⁻³
[OH⁻] = (10⁻³)¹⁽²
[OH⁻] = 0.0316 M
Calculate the pOH
pOH = -log [OH⁻]
pOH = -log [0.0316]
pOH = 1.5
Now calculate pH
pH = 14 - pOH
pH = 14 - 1.5
pH = 12.5
You can also learn about molarity from the following question:
brainly.com/question/14782315
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Answer:
0.07 g/s.
Explanation:
From the question given above, the following data were obtained:
Mass lost = 9.85 g
Time taken = 2 min 30 s
Mean rate =?
Next, we shall convert 2 min 30 s to seconds (s). This can be obtained as follow:
1 min = 60 s
Thus,
2 min = 2 × 60 = 120 s
Therefore,
2 min 30 s = 120 s + 30 s = 150 s
Finally, we shall determine the mean rate of the reaction. This can be obtained as illustrated below:
Mass lost = 9.85 g
Time taken = 150 s
Mean rate =?
Mean rate = mass lost / time taken
Mean rate = 9.85 / 150
Mean rate = 0.07 g/s
Therefore, the mean rate of the reaction is 0.07 g/s
