A 2 x 10^-3 M solution of MnSO4 is gradually made more basic by adding NaOH. At what pH will manganese(II) hydroxide begin to precipitate? [Ksp of Mn(OH)2=2 x 10^-13]
Mn(OH)2 ==> Mn^+2 + 2OH^-
Ksp = (Mn^+2)(OH^-)^2
You know Ksp and Mn, calculate OH^- and from there pOH, then pH. Post your work if you get stuck.
To determine the pH at which manganese(II) hydroxide (Mn(OH)2) begins to precipitate, we need to find the point when the concentration product of Mn(II) and OH- ions exceeds the solubility product constant (Ksp) of Mn(OH)2.
First, let's write the chemical equation for the dissociation of MnSO4 in water:
MnSO4 → Mn2+ + SO42-
Next, we need to consider the reaction between the Mn2+ ions and OH- ions from NaOH:
Mn2+ + 2OH- → Mn(OH)2
Since the initial concentration of MnSO4 is 2 x 10^-3 M and we gradually make the solution more basic by adding NaOH, we can assume that all Mn2+ ions come from the MnSO4 and react with OH- ions.
Let's define the concentration of OH- as x M. Since the reaction stoichiometry is 1:2 between Mn2+ and OH-, the concentration of Mn2+ that reacts with OH- is also x M.
The expression for the solubility product constant of Mn(OH)2 is:
Ksp = [Mn2+][OH-]^2
Substituting the concentrations into the above expression, we get:
2 x 10^-13 = (x)(x)^2
2 x 10^-13 = x^3
Now we need to solve this cubic equation for x, which represents the concentration of OH- ions. We can solve it numerically using a calculator or software that can solve cubic equations.
Once we find the value of x, we can calculate the pOH using the equation:
pOH = -log10([OH-])
Finally, we can calculate the pH using:
pH = 14 - pOH
The pH at which manganese(II) hydroxide begins to precipitate is the pH that corresponds to the concentration of OH- we found.