Respuesta :
Answer: The molar solubility of [tex]Mn(OH)_2[/tex] in a solution that is buffered at ph 8.00 is 0.19 M
Explanation:
Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as [tex]K_{sp}[/tex]
We are given:
Solubility product of [tex]Mn(OH)_2[/tex] = [tex]1.9\times 10^{-13}[/tex]
The equation for the ionization of the [tex]Mn(OH)_2[/tex] is given as:
[tex]Mn(OH)_2\leftrightharpoons Mn^{2+}+2OH^{-}[/tex]
1 mole of [tex]Mn(OH)_2[/tex] gives 1 mole of [tex]Mn^{2+}[/tex] and 2 moles of [tex]OH^{-}[/tex]
Given : pH = 8.00
[tex]pH+pOH=14[/tex]
[tex]pOH=14.0-8.00=6.00[/tex]
[tex]6.00=-log[OH^-][/tex]
[tex][OH^-]=10^{-6}M[/tex]
[tex]K_{sp}=[Mn^{2+}][OH^{-}]^2[/tex]
[tex]1.9\times 10^{-13}M=[s][(10^{-6})^2][/tex]
[tex]s=0.19M[/tex]
Thus the molar solubility of [tex]Mn(OH)_2[/tex] in a solution that is buffered at ph 8.00 is 0.19 M
