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
To calculate the solubility of Mn(OH)2 in grams per liter at different pH levels, we need to consider the dissociation reaction of Mn(OH)2.
The dissociation reaction of Mn(OH)2 can be represented as:
Mn(OH)2(s) ⇌ Mn2+(aq) + 2OH-(aq)
At equilibrium, the solubility product constant (Ksp) expression can be written as:
Ksp = [Mn2+][OH-]^2
The pH of the solution is related to the concentration of hydroxide ions ([OH-]) in the solution. To determine the solubility at different pH levels, we need to find the concentration of hydroxide ions.
Let's break down the calculations for each pH:
a) pH 7.4:
In a neutral solution (pH 7), we have equal concentration of H+ and OH-. Therefore, we can assume the concentration of hydroxide ions ([OH-]) to be 10^(-7.4) M.
Using the Ksp expression, we have: Ksp = [Mn2+][OH-]^2
Since the concentration of OH- is twice the concentration of Mn2+ (according to the balanced equation), let's assume the concentration of Mn2+ to be x.
Ksp = (x)(2[OH-])^2
Substituting the value for [OH-]: Ksp = (x)(2(10^(-7.4))^2
Solve for x (concentration of Mn2+), and this will give you the solubility of Mn(OH)2 in moles per liter (M). Multiply this value by the molar mass of Mn(OH)2 to get the solubility in grams per liter (g/L).
b) pH 9.2:
In a basic solution (pH 9), we need to find the concentration of hydroxide ions ([OH-]). Using the same logic as in part a), the concentration of OH- is 10^(-9.2) M.
Set up and solve the Ksp expression to find the concentration of Mn2+, and then convert it to grams per liter.
c) pH 11.5:
Again, in a basic solution (pH 11.5), we find the concentration of OH- to be 10^(-11.5) M.
Set up and solve the Ksp expression to find the concentration of Mn2+, and then convert it to grams per liter.
Remember to convert moles per liter (M) to grams per liter (g/L) by multiplying the molar mass of Mn(OH)2.
Note: The molar mass of Mn(OH)2 can be calculated by adding the atomic masses of manganese (Mn), hydrogen (H), and oxygen (O) in one molecule of Mn(OH)2.
Let x = solubility Mn(OH)2
Mn(OH)2 ==> Mn^2+ + 2OH^-
...x.........x.......2x
If pH = 7.4. Convert to pH, then to OH^-
Subsitute OH into Ksp for Mn(OH)2 and solve for Mn. That will be x and it will be in M or moles/L. Then convert to grams.
b and c are done the same way.