A solution is 0.20 M in Cd2+ and 0.20 M in Ca2+. NaOH solution is added slowly through a burette.
Ca(OH)2 (s) <------> Ca2+ (aq) +2OH (aq) Ksp= 5.5*10^-6
Cd(OH)2 (s) <-----> Cd2+ (aq) +2OH (aq) Ksp= 2.5*10^-14
Calculate the concentrations of hydroxide ion, calcium ion, and cadmium ion at the point of maximum separation in a selective precipitation.
To calculate the concentrations of hydroxide ion, calcium ion, and cadmium ion at the point of maximum separation in a selective precipitation, we need to consider the solubility product constant (Ksp) and the chemical reactions involved.
1. Let's start by calculating the concentration of hydroxide ion (OH-) at the point of maximum separation.
- At equilibrium, after adding NaOH, the concentration of OH- is the same for both reactions.
- Since 2 moles of OH- are produced for each mole of Ca(OH)2 and Cd(OH)2 that dissolve, the concentration of OH- can be represented as [OH-] = 2x.
- Now, we can write the equilibrium expression for the precipitation of Ca(OH)2: Ksp = [Ca2+][OH-]^2. For Ca(OH)2, Ksp = 5.5 x 10^-6.
- Similarly, the equilibrium expression for the precipitation of Cd(OH)2 is Ksp = [Cd2+][OH-]^2, and for Cd(OH)2, Ksp = 2.5 x 10^-14.
2. Since both Ca(OH)2 and Cd(OH)2 are being precipitated, the OH- concentration is at its maximum when one of the two compounds has fully precipitated.
- To determine this, we compare the Ksp values: Ksp for Ca(OH)2 is higher than that for Cd(OH)2.
- This means that Ca(OH)2 will precipitate first until it reaches its solubility limit, and Cd(OH)2 will remain dissolved.
- At the point of maximum separation, all the Ca2+ ions are precipitated, leaving only Cd2+ ions in solution.
3. Now, we can calculate the concentration of OH- at the point of maximum separation.
- Since Ca(OH)2 has fully precipitated and all Ca2+ ions are removed from the solution, [Ca2+] = 0 M.
- Using the equilibrium expression for Ca(OH)2: Ksp = [Ca2+][OH-]^2 = 5.5 x 10^-6, and substituting [Ca2+] = 0, we can solve for [OH-].
- [OH-]^2 = 5.5 x 10^-6, [OH-] = √(5.5 x 10^-6).
4. Now that we have the concentration of OH-, we can calculate the concentrations of Ca2+ and Cd2+ at the point of maximum separation.
- [Ca2+] = 0 M, as it fully precipitates and no longer remains in solution.
- [Cd2+] remains the same as the initial concentration, which is 0.20 M.
To summarize:
- The concentration of hydroxide ion (OH-) at the point of maximum separation is √(5.5 x 10^-6).
- The concentration of calcium ion (Ca2+) is 0 M.
- The concentration of cadmium ion (Cd2+) is 0.20 M.