How many moles of Fe(OH)2 will dissolve in 1.00L of water buffered at pH = 12.00? Fe(OH)2 has Ksp = 1.8 � 10^-15.
How would I do this? Should I make an ice table? I know the pOH = 2.00.
To determine the number of moles of Fe(OH)2 that will dissolve in 1.00L of water buffered at pH = 12.00, you can use the solubility product constant (Ksp) along with the given information.
First, let's find the concentration of OH- ions in the solution using the pH and pOH relationship. Given that the pH = 12.00, we can calculate the pOH as follows:
pOH = 14 - pH = 14 - 12.00 = 2.00
Next, convert the pOH to OH- concentration using the pOH formula:
pOH = -log[OH-]
10^-pOH = [OH-]
10^-2.00 = 0.01 M (concentration of OH- ions)
Since Fe(OH)2 dissociates into Fe^2+ and 2OH- ions, and the stoichiometry is 1:2, we can assume that 2x moles of Fe(OH)2 will dissociate and give a concentration of 0.01 M OH- ions.
Now, let's solve for x (the concentration of Fe(OH)2) using the Ksp expression:
Ksp = [Fe^2+][OH-]^2
1.8 × 10^-15 = x * (0.01)^2
Rearranging the equation and solving for x:
x = (1.8 × 10^-15) / (0.01)^2
x = 1.8 × 10^-15 / 0.0001
x = 1.8 × 10^-11 moles/L
Therefore, approximately 1.8 × 10^-11 moles of Fe(OH)2 will dissolve in 1.00L of water buffered at pH = 12.00.
You do not need to construct an ice table in this case since you are given the concentration of OH- ions directly.
To find the number of moles of Fe(OH)2 that will dissolve in 1.00L of water buffered at pH = 12.00, you can start by using the relation between pH and pOH.
Given that pOH = 2.00, you can find the concentration of OH- ions using the formula:
pOH = -log[OH-]
Therefore, [OH-] = 10^(-pOH) = 10^(-2) = 0.01 M.
Next, use the stoichiometry of the Fe(OH)2 compound in the balanced equation to determine the concentration of Fe2+ ions. Fe(OH)2 dissociates into 1 Fe2+ ion and 2 OH- ions.
Since the concentration of OH- ions is 0.01 M, the concentration of Fe2+ ions will be half of that, i.e., 0.005 M.
Now, using the solubility product constant (Ksp) value of Fe(OH)2, you can set up an equilibrium expression:
Ksp = [Fe2+][OH-]^2
Substituting the values, we have:
1.8 × 10^(-15) = (0.005)(0.01)^2
Solving this equation will give you the number of moles of Fe(OH)2 that will dissolve in 1.00L of water buffered at pH = 12.00.
Ksp = (Fe^2+)(OH^-)^2.
If pOH is 2 then OH^- is 10^-2 or 0.01. Plug that in, solve for (Fe^2+) and that will be the solubility in mols/L. You have 1 L of solution so that is the # mols.