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
#SPJ4
So you need to find the volume in L? If so:
Convert the mass of Lithium Bromide into moles by dividing the 100 grams by the molar mass of LiBr, taken from the periodic table
In a solution, moles = (concentration in mole/L) x (volume in L)
We know the moles, we have the concentration in mole/L, now find the volume in L, and you should get 0.288. Plz do the math and check for yourself
Density * Volume = Mass
Now we substitute the values in.
19.3 g/cm^3 + 20 cm^3 = 386 g
Mass = 386 g
Answer:
3.5
Explanation:
500*0.175= 8.75 *40/1000=3500/1000=3.5
