Respuesta :
Answer:
[tex]1.29x10^-^6[/tex]
Explanation:
We have to start with the ionization reaction of [tex]Ca(OH)2[/tex] so:
[tex]Ca(OH)_2~-->~Ca^+^2~+~2OH^-^1[/tex] Reaction 1
With this equation we can write the Ksp expression:
[tex]Ksp~=~[Ca^+^2][OH^-^1]^2[/tex]
So, if we want to calculate the Ksp of we have to know the [tex]OH^-^1[/tex] and [tex]Ca^+^2[/tex] concentrations. These concentrations can be calculate using the titration procedure. So, we have to start with the ionic reaction between HCl and Ca(OH)2, so:
[tex]HCl~->~H^+~+~Cl^-[/tex] Reaction 2
[tex]Ca(OH)_2~->~2OH^-~+~Ca^+^2[/tex] Reaction 3
[tex]H^+~+~OH^---->~H_2O[/tex] Reaction4
The volume of HCl used in the experiment is the substraction between the inital and final reading:
[tex]30.23~mL~-~2.77~mL~=~27.46~mL[/tex]
With this volume (27.46 mL = 0.02746 L) we can calculate the moles with the molarity equation, so:
[tex]M=\frac{mol}{L}[/tex]
[tex]0.01~M=\frac{mol}{0.02746~L}[/tex]
[tex]mol~=~0.01*0.02746~=~0.0002746~mol~of~HCl[/tex]
The molar ratio between H+ and OH- is 1:1, therefore:
[tex]0.0002746~mol~of~HCl=0.0002746~mol~of~OH^-[/tex]
Finally we have to calculate the concentration of OH- using the volume (20 mL= 0.02 L) and moles of OH-
[tex]M=\frac{0.0002746~mol~of~OH^-}{0.02~L}[/tex]
[tex]M=~0.01373~M~of~OH^-[/tex]
We already have the concentration of OH-the next step is the calculation of the concentration of [tex]Ca^+^2[/tex]. To do this, we have to use the molar ratio between [tex]Ca^+^2[/tex] and [tex]OH^-[/tex] (reaction 1), the molar ratio is 1:2 therefore:
[tex]0.01373~M~of~OH^-\frac{1}{2}=~0.006865~M~of~Ca^+^2[/tex]
With these values now we can calculate the Ksp:
[tex]Ksp=~[0.006865][0.01373]^2~=~1.29x10^-^6[/tex]
The Ksp value for the [tex]Ca(OH)_2[/tex] is [tex]1.29x10^-^6[/tex]
