Will cadmium hydroxide precipitate from 0.01M solution of cadmium chloride at pH 9. Ksp of cadmium hydroxide =2. 5×10-14 mole2/dm6
pH = 9 means pOH = 5 so (OH^-) = 1E-5
Substitute and solve for Qsp.
Qsp = (Cd^2+)(OH^-)^2 and compare that with Ksp. If Ksp is smaller there is a ppt. If Ksp is larger there is no ppt.
To determine if cadmium hydroxide will precipitate from a 0.01M solution of cadmium chloride at pH 9, we need to compare the solubility product constant (Ksp) of cadmium hydroxide with the concentration of hydroxide ions (OH-) at pH 9.
At pH 9, we can assume that the concentration of hydroxide ions is equal to the concentration of hydroxide ions in pure water, which is 1x10^-7 M.
The balanced equation for the dissociation of cadmium hydroxide is:
Cd(OH)2 (s) ⇌ Cd2+ (aq) + 2OH- (aq)
The equilibrium expression for Ksp of cadmium hydroxide is:
Ksp = [Cd2+] [OH-]^2
Given that Ksp = 2.5x10^-14 mole^2/dm^6, we can substitute the concentration of hydroxide ions into the Ksp expression to find the concentration of cadmium ions (Cd2+):
2.5x10^-14 = [Cd2+] (1x10^-7)^2
Simplifying the equation, we have:
2.5x10^-14 = [Cd2+] (1x10^-14)
Dividing both sides by (1x10^-14), we find:
[Cd2+] = 2.5
From this calculation, we can see that the concentration of cadmium ions is already 2.5 M, which exceeds the concentration of the cadmium chloride solution (0.01 M). This means that cadmium hydroxide will precipitate from the solution at pH 9.
To determine whether cadmium hydroxide (Cd(OH)2) will precipitate from a solution of cadmium chloride (CdCl2) at pH 9, we need to calculate the concentration of hydroxide ions (OH-) in the solution and compare it to the solubility product constant (Ksp) of cadmium hydroxide.
Here's how we can proceed:
1. Write the balanced chemical equation for the dissolution of cadmium hydroxide:
Cd(OH)2(s) ⇌ Cd2+(aq) + 2OH-(aq)
2. Calculate the concentration of hydroxide ions in the solution using the pH value:
Since the solution is at pH 9, we can assume that it is basic. The concentration of hydroxide ions in a basic solution can be calculated using the formula:
[OH-] = 10^(-pOH)
where pOH = 14 - pH.
In this case, pOH = 14 - 9 = 5.
Therefore, [OH-] = 10^(-5) = 0.00001 M.
3. Compare the concentration of hydroxide ions ([OH-]) with the solubility product constant (Ksp) of cadmium hydroxide:
Ksp = [Cd2+][OH-]^2
Since the concentration of cadmium chloride (CdCl2) is not given, we need to assume complete dissociation of CdCl2 to form Cd2+ ions. Therefore, the concentration of Cd2+ in the solution is equal to the initial concentration of cadmium chloride, which is 0.01 M.
Substituting the values into the Ksp expression:
Ksp = (0.01 M)(0.00001 M)^2
Ksp = 2.5 × 10^(-14) mole^2/dm^6
Since Ksp is the equilibrium constant for the reaction, precipitation of cadmium hydroxide will occur if the ion product ([Cd2+][OH-]^2) exceeds the Ksp value.
In this case, the ion product is:
[Cd2+][OH-]^2 = (0.01 M)(0.00001 M)^2
[Cd2+][OH-]^2 = 1 × 10^(-11) mol^3/dm^6
Since 1 × 10^(-11) is greater than the Ksp value of 2.5 × 10^(-14), cadmium hydroxide will precipitate from the solution.