A 6.6 × 10–4 M solution of MnSO4 is gradually made more basic by adding NaOH. At what pH will manganese(II) hydroxide begin to precipitate? For Mn(OH)2, Ksp = 2.0 × 10–13.
Well, it seems like manganese(II) hydroxide is ready to throw a little precipitation party! Let's see when it'll start.
To figure out the pH at which Mn(OH)2 will begin to precipitate, we need to find the concentration of hydroxide ions (OH-) in the solution. This concentration will be determined by the concentration of NaOH we add, and the equation shows that one mole of Mn(OH)2 is formed from 2 moles of OH-.
We know that the concentration of MnSO4 is 6.6 × 10–4 M, which means that the concentration of Mn2+ is also 6.6 × 10–4 M.
Since the concentration of OH- is twice that of Mn2+, we'll have 2 × 6.6 × 10–4 M OH- ions.
Now, let's determine the solubility product constant (Ksp) of Mn(OH)2, which is given as 2.0 × 10–13. Ksp is the product of the concentration of Mn2+ and OH- ions.
Therefore, (6.6 × 10–4) × (2 × 6.6 × 10–4) = 2.0 × 10–13
Simplifying that equation leads to (2.16 × 10–7)^2 ≈ 2.0 × 10–13
Now, drumroll, please! Based on this Ksp value, we can calculate the OH- concentration at which the precipitation will begin.
Taking the square root of both sides, we get:
2.16 × 10–7 ≈ 1.42 × 10–7
So, at a concentration of approximately 1.42 × 10–7 M OH-, manganese(II) hydroxide will start to throw its own little precipitation party.
As for the pH, OH- ions are the ones responsible for alkaline conditions. So, the pOH at which the precipitation starts is -log10(1.42 × 10–7). Calculating that gives us approximately 6.85.
Finally, to find the pH, we'll subtract the pOH from 14:
14 - 6.85 ≈ 7.15
So, the approximate pH at which manganese(II) hydroxide will begin to precipitate is around 7.15. Time to put on your dancing shoes, Mn(OH)2!
To determine at what pH manganese(II) hydroxide, Mn(OH)2, will begin to precipitate, we need to find the point at which the concentration of hydroxide ions (OH-) exceeds the solubility product constant (Ksp) of Mn(OH)2.
The balanced equation for the dissociation of Mn(OH)2 is:
Mn(OH)2(s) ⇌ Mn2+(aq) + 2OH-(aq)
The Ksp expression for Mn(OH)2 is:
Ksp = [Mn2+][OH-]^2
Given that Ksp = 2.0 × 10^-13 and the concentration of MnSO4 is 6.6 × 10^-4 M, we can set up an ICE table to calculate the concentration of Mn2+ and OH- ions as the solution becomes more basic.
Initial:
[Mn2+] = 0 M
[OH-] = 0 M
Change:
[Mn2+] = x M
[OH-] = 2x M (according to the balanced equation)
Equilibrium:
[Mn2+] = x M
[OH-] = 2x M
Substituting these values into the Ksp expression, we get:
Ksp = (x)(2x)^2
2.0 × 10^-13 = 4x^3
Solving for x:
x^3 = (2.0 × 10^-13) / 4
x^3 = 5.0 × 10^-14
x ≈ 6.3 × 10^-5
Since [OH-] = 2x, the concentration of OH- ions is (2)(6.3 × 10^-5) = 1.26 × 10^-4 M.
Now, we can calculate the pOH, which is the negative logarithm of the OH- concentration:
pOH = -log10(1.26 × 10^-4)
pOH ≈ 3.90
Finally, we can calculate the pH using the relationship between pH and pOH:
pH = 14 - pOH
pH ≈ 14 - 3.90
pH ≈ 10.10
Therefore, manganese(II) hydroxide, Mn(OH)2, will begin to precipitate when the pH reaches approximately 10.10.
To determine the pH at which manganese(II) hydroxide (Mn(OH)2) will begin to precipitate, we need to find the concentration of hydroxide ions (OH-) at that point. Since MnSO4 is a soluble compound, we can assume full dissociation in water, resulting in the following equation:
MnSO4 -> Mn2+ + SO42-
From the balanced equation, we can see that the concentration of Mn2+ ions is equal to the initial concentration of the MnSO4 solution; thus, [Mn2+] = 6.6 × 10^(-4) M.
The solubility product (Ksp) for Mn(OH)2 is given as 2.0 × 10^(-13), which has the following equation:
Mn(OH)2 -> Mn2+ + 2OH-
From this equation, we can observe that the concentration of OH- ions will be twice the concentration of Mn2+ ions; hence [OH-] = 2[Mn2+].
Now, let's set up an equilibrium expression for the reaction of Mn(OH)2 using the concentrations:
Ksp = [Mn2+][OH-]^2
Substituting the values we have, we get:
2.0 × 10^(-13) = (6.6 × 10^(-4))(2[OH-])^2
Now, we can solve for [OH-]:
[OH-] = √[(2.0 × 10^(-13)) / (2(6.6 × 10^(-4)))]
[OH-] ≈ 1.12 × 10^(-5) M
Finally, to find the pH, we can use the fact that the concentration of hydroxide ions (OH-) in a solution is related to the pH by the equation:
pOH = -log[OH-]
pOH = -log(1.12 × 10^(-5))
pOH ≈ 4.95
Since pH + pOH = 14, we can calculate the pH:
pH = 14 - 4.95
pH ≈ 9.05
Therefore, manganese(II) hydroxide (Mn(OH)2) will begin to precipitate when the pH reaches approximately 9.05.
(Mn^+2)(OH^-) = 2.0 x 10^-13
Substitute 6.6 x 10^-4 for Mn^+2 in Ksp and solve for OH^-. Convert to pOH, then to pH.