Respuesta :
Answer:
The solubility product of calcium hydroxide is [tex]5.35\times 10^{-6}[/tex].
Explanation:
To calculate the concentration of acid, we use the equation given y 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 acid which is [tex]HCl[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ca(OH)_2[/tex].
We are given:
[tex]n_1=1\\M_1=?0.0983 M\\V_1=11.21 mL\\n_2=2\\M_2=?\\V_2=50.00 mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.0983 M\times 11.21 mL=2\times M_2\times 50.00 mL\\\\M_2=0.01102 M[/tex]
Molarity of calcium hydroxide solution = 0.01102 M
[tex]Ca(OH)_2\rightleftharpoons Ca^{2+}(aq)+2OH^-(aq)[/tex]
1 mole of calcium hydroxide gives 1 mole of calcium ions and 2 moles of hydroxide ions.
[tex][Ca^{2+}]=1\times 0.01102 M=0.01102 M[/tex]
[tex][OH^{-}]=2\times 0.01102 M=0.02204 M[/tex]
Solubility product is given by:
[tex]K_{sp}=[Ca^{2+}][OH^-]^2[/tex]
[tex]=0.01102 M\times (0.02204 M)^2=5.35\times 10^{-6}[/tex]
The solubility product of calcium hydroxide is [tex]5.35\times 10^{-6}[/tex].
