Dr bob for this question :Calculate the molar solubility of manganese II hydroxide in ph 12.85 solution. The Ksp is 2.1 x 10^-13 but I can't solve for x thanks.
Ksp = (Mn^2+)(OH^-)^2
2.1E-13 = (x)(2x + 0.07)^2
To avoid a quadratic assume 2x + 0.07 = 0.07
2.1E-13 = (x)(0.0708)^2
x = (2.1E-13/(0.0708)^2
etc.
To calculate the molar solubility of manganese II hydroxide (Mn(OH)2) in a pH 12.85 solution, you can use the concept of ionization and the Ksp (solubility product constant) value.
The Ksp expression for manganese II hydroxide is:
Ksp = [Mn2+][OH-]^2
Since the Ksp value is given as 2.1 x 10^-13, we can assume that the concentration of Mn2+ and OH- ions at equilibrium is x M.
In a pH 12.85 solution, the concentration of OH- ions can be determined using the pOH expression:
pOH = 14 - pH
pOH = 14 - 12.85 = 1.15
Now, calculate the concentration of OH-:
[OH-] = 10^(-pOH)
[OH-] = 10^(-1.15)
[OH-] ≈ 7.08 x 10^(-2) M
Since the stoichiometry of manganese II hydroxide is 1:2 (1 Mn2+ ion and 2 OH- ions), the equilibrium concentration of Mn2+ ions is also approximately 7.08 x 10^(-2) M.
To calculate the molar solubility, divide the concentration of Mn2+ ions by the molar mass of manganese II hydroxide (Mn(OH)2):
molar solubility = (7.08 x 10^(-2) M) / (molar mass of Mn(OH)2)
The molar mass of Mn(OH)2 = (55.85 g/mol) + 2((1.01 g/mol) + 16.00 g/mol)
= 88.97 g/mol
molar solubility ≈ (7.08 x 10^(-2) M) / (88.97 g/mol)
Thus, the molar solubility of manganese II hydroxide in a pH 12.85 solution is approximately (7.08 x 10^(-2) M) / (88.97 g/mol).