Question

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 ) .

Answer 1

Answer:

53.18 gL⁻¹

Explanation:

Given that:

$Cu^{2+}_{(aq)}$ $+$ $2NH_{3(aq)}$ $------>$  $[Cu(NH_3)_2]^+_{(aq)}$      ------equation (1)

where;

Formation Constant  $(k_f) =$ $6.3*10^{10}$

However, the Dissociation of $CuBr_{(s)$ yields:

$CuBr_{(s)}$      ⇄    $Cu^{+}_{(aq)}$  $+$ $Br^-_{(aq)}$      -------------- equation (2)

where;

the Solubility Constant $(k_{sp})$  $= 6.3 *10^{-9$

From equation (1);

$(k_f) =$ $\frac{[[Cu(NH_3)_2]^+]}{[Cu^{2+}][NH_3]^{2}}$            ---------  equation (3)

From equation (2)

$(k_{sp})$  $= [Cu^+][Br^-]$           ---------  equation (4)

In $NH_3$, the net reaction for $CuBr_{(s)$ can be illustrated as:

$CuBr_{(s)$   $+$ $2NH_{3(aq)}$  ⇄  $[Cu(NH_3)_2]^+_{(aq)}$  $+ Br^-_{(aq)}$

The equilibrium constant (K) can be written as :

$K=$$\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}$

If we multiply both the numerator and the denominator with  $[Cu^+]$ ; we have:

$K=$$\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}*\frac{[Cu^+]}{[Cu^+]}$

$K=$$\frac{[[Cu(NH_3)_2]^+}{[NH_3]^2[Cu^+]}*{[Cu^+][Br^-]}$

$K = k_f *k_{sp}$

$K= (6.3*10^{10})*(6.3*10^{-9})$

$K= 3.97*10^2$

$K$ ≅ $4.0*10^2$

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

$K=$$\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}$

$4.0*10^2 = \frac{(x)(x)}{(0.76-2x)^2}$

$4.0*10^2 = \frac{(x)^2}{(0.76-2x)^2}$

$4.0*10^2 = (\frac{(x)}{(0.76-2x)})^2$

By finding the square of both sides, we have

$\sqrt {4.0*10^2} = \sqrt {(\frac{(x)}{(0.76-2x)})^2$

$2.0*10 = \frac{x}{(0.76-2x)}$

$20(0.76-2x) =x$

$15.2 -40x=x$

$15.2 = 40x +x$

$15.2 = 41x$

$x = \frac{15.2}{41}$

$x = 0.3707 M$

In gL⁻¹; the solubility of $CuBr_{(s)$ in 0.76 M $NH_3$ solution will be:

$= \frac{0.3707 mole of CuBr}{1L}*\frac{143.45 g}{mole of CuBr}$

=  53.18 gL⁻¹