Question

The solubility of copper(i) chloride is 3.91 mg per 100.0 ml of solution. calculate ksp for cucl (cucl=99.00 g mol-1).

Answer 1

Answer : The value of
`K_(sp)` is
`1.56* 10^(-7)`

Explanation :

First we have to calculate the mass of CuCl in 1 L or 1000 mL solution.

As, 100.0 mL of solution contains 3.91 mg of CuCl

So, 1000 mL of solution contains
`(1000mL)/(100.0mL)* 3.91mg=39.1mg` of CuCl

The mass of CuCl = 39.1 mg = 0.0391 g

conversion used : (1 mg = 0.001 g)

Now we have to calculate the moles of CuCl.

`\text{Moles of }CuCl=\frac{\text{Mass of }CuCl}{\text{Molar mass of }CuCl}`

Molar mass of CuCl = 99.00 g/mol

`\text{Moles of }CuCl=(0.0391g)/(99.00g/mol)=3.95* 10^(-4)mole`

Now we have to calculate the moles of
`Cu^+` and
`Cl^-` ion.

Moles of
`Cu^+` = Moles of
`Cl^-` = Moles of CuCl =
`3.95* 10^(-4)mole`

Thus, the concentration of
`Cu^+` and
`Cl^-` ion in 1 L solution will be:

`[Cu^+]` =
`[Cl^-]` =
`3.95* 10^(-4)M`

Now we have to calculate the value of
`K_(sp)` for CuCl.

The solubility equilibrium reaction will be:

`CuCl\rightleftharpoons Cu^(+)+Cl^(-)`

The expression for solubility constant for this reaction will be,

`K_(sp)=[Cu^(+)][Cl^(-)]`

`K_(sp)=(3.95* 10^(-4))* (3.95* 10^(-4))`

`K_(sp)=1.56* 10^(-7)`

Therefore, the value of
`K_(sp)` is
`1.56* 10^(-7)`

Answer 2

convert Mg to grams
1g =1000mg what about 3.91 Mg
= 3.91mg x 1g/1000mg= 3.91 x10^-3 g
moles= mass/molar mass
that is 3.91 x10^-3g /99 g/mol=3.95 x10^-5moles
concentration= moles / vol in liters

that is 3.95 x10^-5/100 x1000= 3.94 x10^-4M

equation for dissociation of CUCl= CUCl----> CU^+ +Cl^-

Ksp=(CU+)(CI-)
that is (3.95 x10^-4)(3.95 x10^-4)
Ksp= 1.56 x10^-7