What is the sequence by which the cations below will precipitate out of solution as the pH is raised?
(Ksp of: Cu(OH)2=4.8x10-20, Sn(OH)2=3x10-27, Cr(OH)3=1.6x10-30)
a. Cu(OH)2, Sn(OH)2, Cr(OH)3
b. Sn(OH)2, Cu(OH)2, Cr(OH)3
c. Sn(OH)2, Cr(OH)3, Cu(OH)2
d.Cr(OH)3, Sn(OH)2, Cu(OH)2
To answer this question completely, one needs to know the concentration of the cation. Let's say the cation concentration is 0.1 M in each case. I'll do Cu(OH)2 for you. You do the others.
..................Cu(OH)2 ==> Cu^2+ + 2OH^-
Ksp = (Cu^2+)(OH^-) = 4.8E-20. Solve for (OH^-).
(OH^-) = sqrt [Ksp/(Cu^2+)] = sqrt (4.8E-20/0.1) = sqrt (48E-20) = 6.9E-10 M
Calculate (OH^-) similarly for the other two salts, the sequence them by pH. If it will make it easier, convert the (OH^-) in each case to pH.
Post your work if you get stuck.
To determine the sequence by which cations will precipitate out of solution as the pH is raised, we need to compare the solubility product constants (Ksp) of the given cations' hydroxides.
The solubility product constant (Ksp) is an equilibrium constant that relates to the saturation of a compound in a solution. The higher the Ksp value, the more soluble the compound is, and the lower the Ksp value, the less soluble the compound is. In this case, we are comparing the solubility of Cu(OH)2, Sn(OH)2, and Cr(OH)3.
We can compare the Ksp values to determine the relative solubilities of the hydroxides:
1. Cu(OH)2 has a Ksp of 4.8 × 10^(-20)
2. Sn(OH)2 has a Ksp of 3 × 10^(-27)
3. Cr(OH)3 has a Ksp of 1.6 × 10^(-30)
The lower the Ksp value, the less soluble the compound is. Hence, the cation with the lowest Ksp value will precipitate out first as the pH is raised.
Comparing the Ksp values given, we can see that Cr(OH)3 has the lowest Ksp value and will, therefore, precipitate out first.
So, the correct sequence by which the cations will precipitate out of solution as the pH is raised is: d. Cr(OH)3, Sn(OH)2, Cu(OH)2.