Suppose that two hydroxides, MOH and M'(OH)2, both have Ksp = 1.0 10-12 and that initially both cations are present in a solution at concentrations of 0.0036 mol · L-1. Which hydroxide precipitates first?
I know MOH ppt's first...but am not sure how to approach this second part of the problem.
At what pH does the first hydroxide precipitate, when solid NaOH is added?
All you have to do is find the concentration of OH- and solve for pH (pOH= -log [OH-] nad then 14-pOH=ph
[OH-} = Ksp/[M+}
I agree with Trixie. I think the pH is about 4 (note that the ppt is forming WHEN the pH is ACID).
To determine which hydroxide precipitates first, we need to compare their solubility product constants (Ksp) and concentrations. The hydroxide with a smaller Ksp value and a higher concentration will precipitate first.
In this case, both hydroxides have the same Ksp value (1.0 x 10^(-12)). However, to determine which hydroxide precipitates first, we need to compare their concentrations.
Both cations are present in equal concentrations of 0.0036 mol/L, so the concentration factor is the same for both hydroxides. Therefore, we need to consider the effect of the common ion on the solubility.
Since we're adding solid NaOH, it will dissociate in water to form Na+ and OH- ions. The OH- ion is a common ion for the precipitation reaction of both hydroxides. As the concentration of the OH- ion increases, the solubility of both hydroxides will decrease.
However, since M'(OH)2 has two OH- ions per formula unit, it will experience twice the decrease in solubility compared to MOH. Therefore, M'(OH)2 will precipitate first.
Now, to determine the pH at which the first hydroxide (MOH) precipitates when solid NaOH is added, we need to consider the reaction between NaOH and water, which produces OH- ions. The reaction is as follows:
NaOH (s) + H2O (l) → Na+ (aq) + OH- (aq)
At the equivalence point, all the moles of NaOH added have reacted to produce OH- ions. The pH at the equivalence point can be calculated using the following equation:
pOH = -log10 [OH-]
Since the concentration of OH- is known to be 0.0036 mol/L (based on the given initial concentration of the cations), we can calculate the pOH as follows:
pOH = -log10(0.0036)
Using a calculator, pOH ≈ 2.44
Since pOH + pH = 14 (at 25°C), the pH at the equivalence point can be calculated as:
pH = 14 - pOH
pH = 14 - 2.44
pH ≈ 11.56
Therefore, the first hydroxide (MOH) will precipitate when the pH reaches approximately 11.56 when solid NaOH is added.
To determine which hydroxide precipitates first, we need to compare the solubility products, Ksp, of the two hydroxides.
The solubility product expression for MOH is:
Ksp = [M+][OH-]
The solubility product expression for M'(OH)2 is:
Ksp = [M'+]2[OH-]2
Given that both hydroxides have the same Ksp value, which is 1.0 × 10-12, we can write the equilibrium expressions for both hydroxides as:
For MOH:
1.0 × 10-12 = [M+][OH-]
For M'(OH)2:
1.0 × 10-12 = [M'+]2[OH-]2
Since the concentrations of cations (M+ and M'+) are the same at 0.0036 mol/L, we can assume their values are equal in both expressions. Therefore, we can compare the concentrations of hydroxide ions ([OH-]) in each expression to determine which hydroxide precipitates first.
1. For MOH:
Ksp = [M+][OH-]
1.0 × 10-12 = [M+][OH-]
[OH-] = (1.0 × 10-12) / [M+]
2. For M'(OH)2:
Ksp = [M'+]2[OH-]2
1.0 × 10-12 = [M'+]2[OH-]2
[OH-] = sqrt((1.0 × 10-12) / ([M'+]2))
To compare the concentrations of hydroxide ions, we need to find the value of [M+]' and [M'+].
However, the concentrations of the cations are not given in the problem statement, so it is not possible to determine which hydroxide precipitates first based on the information provided.
Regarding the pH at which the first hydroxide precipitates when solid NaOH is added, we know that NaOH dissociates in water to produce Na+ and OH- ions. When solid NaOH is added, OH- ions will be released into the solution, increasing the concentration of OH-.
The pH at which precipitation occurs can be determined by finding the point at which the concentration of hydroxide ions [OH-] exceeds the solubility product (Ksp) of the hydroxide compound. In this case, the hydroxide compound that precipitates first would be the one with the lower Ksp value (assuming no other factors like complexation).
Unfortunately, the pH at which precipitation occurs cannot be determined accurately without additional information about the pKa or pKsp values of these hydroxides or their cations.