Answer : The value of [tex]K_{sp}[/tex] is [tex]5.04\times 10^{-6}[/tex]
Explanation :
To calculate the concentration of [tex]Ca(OH)_2[/tex], we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of [tex]Ca(OH)_2[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of HCl.
We are given:
[tex]n_1=2\\M_1=?\\V_1=50.00mL\\n_2=1\\M_2=0.0973M\\V_2=11.15mL[/tex]
Putting values in above equation, we get:
[tex]2\times M_1\times 50.00mL=1\times 0.0973M\times 11.15mL[/tex]
[tex]M_1=0.0108M[/tex]
Now we have to calculate the [tex]K_{sp}[/tex] for [tex]Ca(OH)_2[/tex]
The solubility equilibrium reaction will be:
[tex]Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^{-}[/tex]
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Ca^{2+}][OH^{-}]^2[/tex]
Let the solubility be, x
x = 0.0108 M
[tex]K_{sp}=(x)\times (2x)^2[/tex]
[tex]K_{sp}=(0.0108)\times (2\times 0.0108)^2[/tex]
[tex]K_{sp}=5.04\times 10^{-6}[/tex]
Therefore, the value of [tex]K_{sp}[/tex] is [tex]5.04\times 10^{-6}[/tex]
