Answer : The pH of a saturated solution is, 12.33
Explanation : Given,
= ![5.02\times 10^{-6}](https://tex.z-dn.net/?f=5.02%5Ctimes%2010%5E%7B-6%7D)
First we have to calculate the solubility of
ion.
The balanced equilibrium reaction will be:
![Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-](https://tex.z-dn.net/?f=Ca%28OH%29_2%5Crightleftharpoons%20Ca%5E%7B2%2B%7D%2B2OH%5E-)
Let the solubility will be, 's'.
The concentration of
ion = s
The concentration of
ion = 2s
The expression for solubility constant for this reaction will be,
![K_{sp}=[Ca^{2+}][OH^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BOH%5E-%5D%5E2)
Let the solubility will be, 's'
![K_{sp}=(s)\times (2s)^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%28s%29%5Ctimes%20%282s%29%5E2)
![K_{sp}=(4s)^3](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%284s%29%5E3)
Now put the value of
in this expression, we get the solubility.
![5.02\times 10^{-6}=(4s)^3](https://tex.z-dn.net/?f=5.02%5Ctimes%2010%5E%7B-6%7D%3D%284s%29%5E3)
![s=1.079\times 10^{-2}M](https://tex.z-dn.net/?f=s%3D1.079%5Ctimes%2010%5E%7B-2%7DM)
The concentration of
ion = s = ![1.079\times 10^{-2}M](https://tex.z-dn.net/?f=1.079%5Ctimes%2010%5E%7B-2%7DM)
The concentration of
ion = 2s = ![2\times (1.079\times 10^{-2}M)=2.158\times 10^{-2}M](https://tex.z-dn.net/?f=2%5Ctimes%20%281.079%5Ctimes%2010%5E%7B-2%7DM%29%3D2.158%5Ctimes%2010%5E%7B-2%7DM)
First we have to calculate the pOH.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)
![pOH=-\log (2.158\times 10^{-2})](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%282.158%5Ctimes%2010%5E%7B-2%7D%29)
![pOH=1.67](https://tex.z-dn.net/?f=pOH%3D1.67)
Now we have to calculate the pH.
![pH+pOH=14\\\\pH=14-pOH\\\\pH=14-1.67=12.33](https://tex.z-dn.net/?f=pH%2BpOH%3D14%5C%5C%5C%5CpH%3D14-pOH%5C%5C%5C%5CpH%3D14-1.67%3D12.33)
Therefore, the pH of a saturated solution is, 12.33