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
See your post above.
To calculate the solubility of Mn(OH)2 in grams per liter at a specific pH, we need to use the concept of the solubility product constant, also known as Ksp. The solubility product constant describes the equilibrium between a solid compound and its dissolved ions in a solution.
The chemical equation for the dissociation of Mn(OH)2 is:
Mn(OH)2 ⇌ Mn2+ + 2OH-
The solubility product constant expression for this equation can be written as:
Ksp = [Mn2+][OH-]^2
To calculate the solubility, we need to find the concentration of Mn2+ ions and OH- ions at a particular pH. We can assume the concentration of OH- ions is equal to the concentration of Mn2+ ions because the ratio is 1:2 in the balanced equation.
Let's calculate the solubility for each pH:
a) pH 7.4:
At pH 7.4, we can assume the solution is buffered and the concentration of OH- ions can be determined using the following equation:
pOH + pH = 14
Since the pH is 7.4, the pOH can be calculated as follows:
pOH = 14 - 7.4 = 6.6
Now, convert pOH to OH- concentration:
[OH-] = 10^(-pOH)
[OH-] = 10^(-6.6) = 2.5 x 10^(-7) M
Since the concentration of OH- ions is equal to the concentration of Mn2+ ions, we can use the solubility product constant expression to find the solubility:
Ksp = [Mn2+][OH-]^2
Ksp = [Mn2+][(2.5 x 10^(-7))^2]
Solving for [Mn2+], we have:
[Mn2+] = Ksp / (2.5 x 10^(-7))^2
Substituting the Ksp value for Mn(OH)2 (you can find this value in a reference source or given in the question), you can calculate the solubility of Mn(OH)2 in grams per liter.
Repeat similar calculations for parts b) and c) using the given pH values to determine the concentration of OH- ions and then finding the solubility of Mn(OH)2 using the expression for Ksp.
Note: The Ksp value for Mn(OH)2 can vary depending on the temperature. Be sure to use the appropriate value for the given temperature conditions.