A mineral sample is analyzed for its cobalt and calcium content. A sample is dissolved, and then the cobalt and calcium are precipitated as Co(OH)2(s) and Ca(OH)2(s). At what pH can Co(OH)2(s) be separated from Ca(OH)2(s) at 25 °C? Assume that an effective separation requires a maximum concentration of the less soluble hydroxide of 1× 10^–6 M.
Please help! So far I have gotten 12.46 as the PH and .0291 as the concentration but it's wrong! Help ! I'm super confused.... Thank you in advance!!
If we are to obtain the same answers we should have the Ksp values you're working with.
These are the given Ksp values to use:
Ksp of Co(OH)2 is 5.9x10^-15 M^3
Ksp of Ca(OH)2 is 5.0x10^-6 M^3
I would do this.
Ksp Co(OH)2 = 5.9E-15 so
(OH^-) = sqrt(Ksp/1E-6)
Then convert OH to pH. I get an answer of something like 9.9 or so.
Thanks Dr.Bob!! I was then able to plug into the second equation and found the concentration of calcium! You were great help! Thanks again :)
To determine the pH at which Co(OH)2(s) can be separated from Ca(OH)2(s) at 25 °C, we need to consider the solubility products of both hydroxides.
Let's start by writing the solubility product expressions for Co(OH)2(s) and Ca(OH)2(s):
Co(OH)2(s) ⇌ Co2+(aq) + 2OH^-(aq)
Ksp(Co(OH)2) = [Co2+][OH^-]^2
Ca(OH)2(s) ⇌ Ca2+(aq) + 2OH^-(aq)
Ksp(Ca(OH)2) = [Ca2+][OH^-]^2
For effective separation, we want to achieve a maximum concentration of 1× 10^–6 M of the less soluble hydroxide. In this case, it is Co(OH)2. So, we need to find the pH at which [OH^-] = 1× 10^–6 M for Co(OH)2 while maintaining a lower concentration for Ca(OH)2.
To determine the pH, we can use the fact that at 25 °C, the equilibrium constant Kw (the product of the concentrations of hydrogen ions [H+] and hydroxide ions [OH^-]) is equal to 1.0 × 10^-14.
Kw = [H+][OH^-] = 1.0 × 10^-14
Since [OH^-] is already given as 1× 10^-6 M, we can solve for [H+] by dividing Kw by [OH^-]:
[H+] = Kw / [OH^-] = (1.0 × 10^-14) / (1.0 × 10^-6) = 1.0 × 10^-8 M
To find the pH, we take the negative logarithm of [H+]:
pH = -log[H+] = -log(1.0 × 10^-8) = 8
Therefore, at a pH of 8, Co(OH)2(s) can be separated from Ca(OH)2(s) at 25 °C.
It seems there might be an error in your previous calculations. The correct pH for the separation is 8, not 12.46.