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.
Ksp[Co(OH02]=5.9e-15
Ksp[Ca(OH02]=5.0e-6
What is the concentration of the more soluble hydroxide at this pH?
I got 9.89 for the pH, and that was right. I am having trouble with the concentration.
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457/854=78mol
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To determine the concentration of the more soluble hydroxide at the given pH, we need to compare the solubility product constants (Ksp) of Co(OH)2 and Ca(OH)2.
Let's start by writing the dissociation equations for these compounds:
Co(OH)2(s) ⇌ Co2+(aq) + 2OH-(aq)
Ca(OH)2(s) ⇌ Ca2+(aq) + 2OH-(aq)
The solubility product constant for Co(OH)2 is given as Ksp[Co(OH)2] = 5.9e-15, which can be expressed as follows:
Ksp[Co(OH)2] = [Co2+][OH-]^2
Since we assume an effective separation requires a maximum concentration of 1×10^-6 M for the less soluble hydroxide, we can set [OH-] = 1×10^-6 M.
Now, we need to find the concentration of Co2+. Rearranging the equation, we have:
[Co2+] = Ksp[Co(OH)2] / [OH-]^2
= (5.9e-15) / (1e-6)^2
= 5.9e-15 / 1e-12
= 5.9e-3 M
So, the concentration of Co2+ at this pH is 5.9e-3 M.
Now, let's look at the solubility product constant for Ca(OH)2, given as Ksp[Ca(OH)2] = 5.0e-6:
Ksp[Ca(OH)2] = [Ca2+][OH-]^2
Since we already determined that [OH-] = 1×10^-6 M, we can rearrange the equation to find [Ca2+]:
[Ca2+] = Ksp[Ca(OH)2] / [OH-]^2
= (5.0e-6) / (1e-6)^2
= 5.0e-6 / 1e-12
= 5.0e-6 M
Therefore, the concentration of Ca2+ at this pH is 5.0e-6 M.
In summary, at the given pH of 9.89, the concentration of the more soluble hydroxide (Co2+) is 5.9e-3 M, and the concentration of the less soluble hydroxide (Ca2+) is 5.0e-6 M.