First we must express the Ksp expression for
C2D3 to be in terms of molar solubility x.
C2D3 = (2x)^2 * (3x)^3= 108 x^5
Then we set that equation to be equal to our solubility
constant
9.14 x 10^-9 = 108 x^5
Calculate for x:
x = 9.67 x 10^-3
So the molar solubility is 9.67 x 10^-3
Answer: [tex]9.67\times 10^{-3}[/tex] moles /liter
Explanation: The equation for the reaction will be as follows:
[tex]C_2d_3\leftrightharpoons 2C^{3+}+3d^{2-}[/tex]
1 mole of [tex]C_2d_3[/tex] gives 2 moles of [tex]C^{3+}[/tex] and 3 moles of [tex]d^{2-}[/tex].
Thus if solubility of [tex]C_2d_3[/tex] is s moles/liter, solubility of [tex]C^{3+}[/tex] is 2s moles\liter and solubility of [tex]d^{2-}[/tex] is 3s moles/liter
Therefore,
[tex]K_sp=[C^{3+}]^2[d^{2-}^3][/tex]
[tex]9.14\times 10^{-9}=[2s]^2[3s]^3[/tex]
[tex]108s^5=9.14\times 10^{-9}[/tex]
[tex]s=9.67\times 10^{-3}moles/liter[/tex]
