Copper(I) ions in aqueous solution react with NH 3 ( aq ) according to Cu + ( aq ) + 2 NH 3 ( aq ) ⟶ Cu ( NH 3 ) + 2 ( aq ) K f = 6.3 × 10 10 Calculate the solubility (in g·L−1) of CuBr ( s ) ( K sp = 6.3 × 10 − 9 ) in 0.76 M NH 3 ( aq ) .

53.18 gL⁻¹

Explanation:

Given that:

       ------equation (1)

where;

Formation Constant  

However, the Dissociation of yields:

     ⇄            -------------- equation (2)

where;

the Solubility Constant  

From equation (1);

           ---------  equation (3)

From equation (2)

            ---------  equation (4)

In , the net reaction for can be illustrated as:

   ⇄    

The equilibrium constant (K) can be written as :

If we multiply both the numerator and the denominator with   ; we have:

≅

Now; we can re-write our equilibrium constant again as:

By finding the square of both sides, we have

In gL⁻¹; the solubility of in 0.76 M solution will be:

=  53.18 gL⁻¹