Respuesta :
Answer: The [tex]K_{sp}[/tex] value for calcium hydroxide at this temperature is [tex]1.08 \times 10^{-4}[/tex].
Explanation:
Given: Mass of [tex]Ca(OH)_{2}[/tex] = 0.225 g
Volume = 0.100 L
As moles is the mass of substance divided by its molar mass.
So, moles of [tex]Ca(OH)_{2}[/tex] (molar mass = 74 g/mol) is calculated as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{0.225 g}{74 g/mol}\\= 0.003 mol[/tex]
Molarity is the number of moles of substance present in a liter of solution.
Hence, molarity of given solution will be as follows.
[tex]Molarity = \frac{moles}{Volume (in L)}\\= \frac{0.003 mol}{0.1 L}\\= 0.03 M[/tex]
The equation for dissociation of [tex]Ca(OH)_{2}[/tex] is as follows.
[tex]Ca(OH)_{2} \rightarrow Ca^{2+} + 2OH^{-}[/tex]
This means that [tex][Ca^{2+}] = 0.03[/tex] and [tex][OH^{-}] = 2 \times 0.03 = 0.06[/tex]. Hence, [tex]K_{sp}[/tex] value for this reaction is calculated as follows.
[tex]K_{sp} = [Ca^{2+}][OH^{-}]^{2}\\= (0.03) \times (0.06)^{2}\\= 1.08 \times 10^{-4}[/tex]
Thus, we can conclude that the [tex]K_{sp}[/tex] value for calcium hydroxide at this temperature is [tex]1.08 \times 10^{-4}[/tex].
