Respuesta :
Answer : The pH of a saturated solution is, 12.33
Explanation : Given,
[tex]K_{sp}[/tex] = [tex]5.02\times 10^{-6}[/tex]
First we have to calculate the solubility of [tex]OH^-[/tex] ion.
The balanced equilibrium reaction will be:
[tex]Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-[/tex]
Let the solubility will be, 's'.
The concentration of [tex]Ca^{2+}[/tex] ion = s
The concentration of [tex]OH^-[/tex] ion = 2s
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Ca^{2+}][OH^-]^2[/tex]
Let the solubility will be, 's'
[tex]K_{sp}=(s)\times (2s)^2[/tex]
[tex]K_{sp}=(4s)^3[/tex]
Now put the value of [tex]K_[sp}[/tex] in this expression, we get the solubility.
[tex]5.02\times 10^{-6}=(4s)^3[/tex]
[tex]s=1.079\times 10^{-2}M[/tex]
The concentration of [tex]Ca^{2+}[/tex] ion = s = [tex]1.079\times 10^{-2}M[/tex]
The concentration of [tex]OH^-[/tex] ion = 2s = [tex]2\times (1.079\times 10^{-2}M)=2.158\times 10^{-2}M[/tex]
First we have to calculate the pOH.
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log (2.158\times 10^{-2})[/tex]
[tex]pOH=1.67[/tex]
Now we have to calculate the pH.
[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-1.67=12.33[/tex]
Therefore, the pH of a saturated solution is, 12.33
