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

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

Answer 2

Answer : The value of $K_{sp}$ is $1.56\times 10^{-7}$

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 $\frac{1000mL}{100.0mL}\times 3.91mg=39.1mg$ 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

$\text{Moles of }CuCl=\frac{0.0391g}{99.00g/mol}=3.95\times 10^{-4}mole$

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

Moles of $Cu^+$ = Moles of $Cl^-$ = Moles of CuCl = $3.95\times 10^{-4}mole$

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

$[Cu^+]$ = $[Cl^-]$ = $3.95\times 10^{-4}M$

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,

$K_{sp}=[Cu^{+}][Cl^{-}]$

$K_{sp}=(3.95\times 10^{-4})\times (3.95\times 10^{-4})$

$K_{sp}=1.56\times 10^{-7}$

Therefore, the value of $K_{sp}$ is $1.56\times 10^{-7}$