Question

Copper(I) ions in aqueous solution react with NH3(aq) according to Cut(aq)2 NH (aq) - Cu(NH(aq) = 6.3 x 1010 Calculate the solubility (in g-L) of CuBr(s) (Ksp = 6.3 x 109) in 0.64 M NH3 (aq) solubility of CuBr(s): g/L

Answer 1

Consider reaction, CuBr (s) Cu ⁺ (aq) + Br ^- (aq)   (1)

For above reaction, equilibrium constant is K sp = [ Cu ⁺ ] [Br ^- ] = 6.3 10 ^-09

When CuBr is added to solution containing ammonia , a complex formation reaction takes place due to which solubility of CuBr increases. Consider a complex formation reaction as shown below.

Cu ⁺ (aq) + 2 NH ₃ (aq)   [ Cu (NH ₃ ) ₂ ] ⁺ (aq)    (2)

For above reaction, equilibrium constant is K f = [ Cu (NH ₃ ) ₂ ] ⁺ / [  Cu ⁺ ] [ NH ₃ ] ² = 6.3 10 ¹⁰

Now , adding equation 1 and 2 , we get equation for dissolution of CuBr in the presence of NH ₃

CuBr (s) Cu + (aq) + Br - (aq) + Cu + (aq) + 2 NH 3 (aq)   [ Cu (NH 3 ) 2 ] + (aq)

CuBr (s) + 2 NH 3 (aq)   [ Cu (NH 3 ) 2 ] + (aq) + Br - (aq)

For overall reaction, equilibrium constant K = K sp K f = 6.3 10 ^-09 6.3 10 ¹⁰ = 3.97 10 ²

Let's use ICE table.

Concentration (M)	CuBr (s)	2 NH 3	[ Cu (NH 3 ) 2 ] +	Br -
I		0.64	0	0
C		- 2 A	+A	+A
E		0.64 - 2 A	A	A

From above values , we can write

K f = ( A) ( A) / ( 0.64 - 2 A ) ² = 3.97 10 ²

$\therefore$ A ² / ( 0.64 - 2 A ) ² = 3.97 10 ²

$\therefore$ A / 0.64 - 2 A = 19.92

A = 19.92 ( 0.64 -2 A )

A = 12.75 - 39.84 A

A + 39.84 A = 12.75

40.84 A = 12.75

A = 12.75 / 40.84 = 0.31 M = [ Br ^- ] = Solubility of CuBr in ammonia solution  

Molar mass of CuBr = 63.54 + 79.91 = 143.45 g / mol

Solubility of CuBr in g / L = Solubility of CuBr in mol / L Molar mass of CuBr

Solubility of CuBr in g / L = 0.31 mol / L 143.45 g / mol = 44.46 g / L

ANSWER : Solubility of CuBr in 0.64 M ammonia solution = 44 g / L