Question

C2d3 has a solubility product constant of 9.14×10−9. what is the molar solubility of c2d3? express your answer with the appropriate units.

Answer 1

Answer:
`9.67* 10^(-3)` moles /liter

Step-by-step explanation: The equation for the reaction will be as follows:

`C_2d_3\leftrightharpoons 2C^(3+)+3d^(2-)`

1 mole of
`C_2d_3` gives 2 moles of
`C^(3+)` and 3 moles of
`d^(2-)`.

Thus if solubility of
`C_2d_3` is s moles/liter, solubility of
`C^(3+)` is 2s moles\liter and solubility of
`d^(2-)` is 3s moles/liter

Therefore,

`K_sp=[C^(3+)]^2[d^(2-)^3]`

`9.14* 10^(-9)=[2s]^2[3s]^3`

`108s^5=9.14* 10^(-9)`

`s=9.67* 10^(-3)moles/liter`

Answer 2

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