Calculate the molar solubility of Mn(OH)2 in grams per liter when buffered at pH 8.5 Mn(OH)2=88.953g/mol; Ksp=1.9 x 10^-13
To calculate the molar solubility of Mn(OH)2 in grams per liter, we need to use the solubility product constant (Ksp) and the pH of the solution.
Step 1: Write the balanced equation for the dissociation of Mn(OH)2.
Mn(OH)2(s) ⇌ Mn^2+(aq) + 2OH^-(aq)
Step 2: Write the expression for the solubility product constant (Ksp) using the balanced equation.
Ksp = [Mn^2+][OH^-]^2
Step 3: Convert the pH of the solution to the hydroxide ion concentration [OH^-].
Since the solution is buffered at pH 8.5, we can use the equation pOH = 14 - pH to calculate pOH, and then convert it to [OH^-] using the equation pOH = -log [OH^-].
pOH = 14 - pH = 14 - 8.5 = 5.5
[OH^-] = 10^(-pOH) = 10^(-5.5)
Step 4: Use the solubility product constant (Ksp) expression to determine the concentration of Mn^2+.
Ksp = [Mn^2+][OH^-]^2
1.9 x 10^(-13) = [Mn^2+][10^(-5.5)]^2
1.9 x 10^(-13) = [Mn^2+][10^(-11)]
[Mn^2+] = (1.9 x 10^(-13)) / (10^(-11)) = 1.9 x 10^(-2) M
Step 5: Calculate the molar solubility of Mn(OH)2 in moles per liter.
Since the stoichiometry of the reaction is 1:1, the molar solubility of Mn(OH)2 is equal to the concentration of Mn^2+.
Molar solubility = [Mn^2+] = 1.9 x 10^(-2) M
Step 6: Convert the molar solubility to grams per liter.
Molar solubility (in moles per liter) × molar mass = grams per liter
Molar solubility = 1.9 x 10^(-2) M
Molar mass of Mn(OH)2 = 88.953 g/mol
Molar solubility (in grams per liter) = (1.9 x 10^(-2) M) × (88.953 g/mol) = 1.68 x 10^(-1) g/L
Therefore, the molar solubility of Mn(OH)2 in grams per liter when buffered at pH 8.5 is approximately 1.68 x 10^(-1) g/L.