Calculate the solubility of Mn(OH)2 in grams per liter when buffered at each of the following.

a) pH 7.4

b) pH 9.2

c) pH 11.5

ALL IN GRAMS PER LITER.
PLEASE HELP ME AND EXPLAIN

To calculate the solubility of Mn(OH)2 in grams per liter at different pH values, we need to consider the equilibrium of the reaction:

Mn(OH)2 (s) ⇌ Mn2+ (aq) + 2OH- (aq)

The solubility of Mn(OH)2 depends on the concentration of Mn2+ and OH- ions in solution. The pH of the solution determines the concentration of OH- ions.

a) pH 7.4:
At pH 7.4, the concentration of OH- ions can be calculated using the equilibrium of water:

H2O (l) ⇌ H+ (aq) + OH- (aq)

In pure water, the concentration of H+ ions and OH- ions are equal at 1 x 10^-7 M. However, in a buffered solution at pH 7.4, the concentration of H+ ions is different.

The concentration of OH- ions can be calculated using the equation:

pOH = 14 - pH

pOH = 14 - 7.4

pOH = 6.6

[OH-] = 10^(-pOH)

[OH-] = 10^(-6.6)

[OH-] = 2.51 × 10^(-7) M

Now, we can calculate the solubility of Mn(OH)2 using the Ksp expression:

Ksp = [Mn2+] [OH-]^2

Assuming that the dissociation of Mn(OH)2 is complete and equilibrium concentrations are equal to each other:

[Mn2+] = x (solubility)
[OH-] = 2x (solubility)

Ksp = x * (2x)^2

Ksp = 4x^3

At equilibrium, Ksp = solubility^3

solubility^3 = Ksp

solubility = Ksp^(1/3)

Substituting the known values:

solubility = (4 * (2.51 × 10^(-7))^3)^(1/3)

solubility = 6.79 × 10^(-3) grams per liter

Therefore, the solubility of Mn(OH)2 in a solution buffered at pH 7.4 is approximately 6.79 × 10^(-3) grams per liter.

You can follow a similar procedure for parts b) and c) to calculate the solubility at pH 9.2 and pH 11.5 by using the corresponding pOH values.

b) pH 9.2:
Calculate [OH-] using pOH = 14 - pH and then determine solubility using the Ksp expression.

c) pH 11.5:
Calculate [OH-] using pOH = 14 - pH and then determine solubility using the Ksp expression.

To calculate the solubility of Mn(OH)2 in grams per liter at different pH values, we need to consider the equilibrium reaction of Mn(OH)2 dissociating in water:

Mn(OH)2 ⇌ Mn2+ + 2OH-

The solubility of a compound depends on its solubility product constant, Ksp, which is the equilibrium constant for the dissociation reaction. The expression for the solubility product constant is:

Ksp = [Mn2+][OH-]²

The hydroxide ion concentration, [OH-], depends on the pH of the solution since it is determined by the auto-ionization of water:

[H2O] ⇌ H+ + OH-

For neutral solutions, the concentration of hydroxide ions is the same as the concentration of hydrogen ions: [OH-] = [H+].

To find the solubility of Mn(OH)2 in grams per liter, we can rearrange the solubility product expression and solve for the concentration of Mn2+ ions, which is equal to the solubility of Mn(OH)2:

[Mn2+] = sqrt(Ksp / [OH-])

Now, let's calculate the solubility at each pH value.

a) pH 7.4:
At pH 7.4, the concentration of hydroxide ions can be calculated using the auto-ionization of water:

[H2O] = 1.0 x 10^-7 M (at 25°C)

[H+] = 10^(-pH)

[OH-] = [H+] = 10^(-7.4) M

Substituting these values into the concentration expression:

[Mn2+] = sqrt(Ksp / [OH-])

Here, we need the value of Ksp for Mn(OH)2, which is 2.0 x 10^-13.

[Mn2+] = sqrt(2.0 x 10^-13 / 10^(-7.4))

Calculating this expression will give you the solubility of Mn(OH)2 in grams per liter at pH 7.4.

b) pH 9.2:
Using the same steps as explained above, calculate the concentration of hydroxide ions at pH 9.2:

[OH-] = [H+] = 10^(-9.2) M

Then substitute this value into the concentration expression and solve for [Mn2+], using the Ksp value for Mn(OH)2.

c) pH 11.5:
Repeat the same process for calculating [OH-] at pH 11.5 and substitute this value into the concentration expression and solve for [Mn2+], using the Ksp value for Mn(OH)2.

By applying these steps, you can calculate the solubility of Mn(OH)2 in grams per liter at each of the specified pH values.

676t

Didn't I do this for you yesterday?

Convert pH to OH^-. I would do that by converting to pOH by using
pH + pOH = pKw = 14, then to OH^- by
pOH = -log (OH^-).
Let x = solubility Mn(OH)2.
........Mn(OH)2 => Mn^2+ + 2OH^-
...........x........x........2x

Ksp = (Mn^2+)(OH^-)^2
Substitute Ksp.
(Mn^+) = x
(OH^-) = x + (OH^-) from above.
Solve for x = Mn(OH)2 in moles/L. Convert to g/L by g = moles x molar mass. The others are done the same way. Post your work if you get stuck.