Question

2. The Ksp of Cu2S is 1.0 x 10-48. What concentration of Cu+ (aq) is needed to begin precipitation of Cu>S(s) from a 2.0 x 10 15 m s%(ag) solution?

Answer 1

Consider reaction , Cu₂S (s) $\rightleftharpoons$ 2 Cu ⁺ (aq) + S ^2- (aq)  

For above reaction, K sp = [ Cu ⁺ ] ² [ S ^2- ] = 1.0 $\times$ 10 ^-48 .

We know that, formation of solid takes place when Q = K sp.

Here , we have values of Q= K sp = 1.0 $\times$ 10 ^-48   &  [ S ^2- ] = 2.0 $\times$ 10 ^-15 . We need to calculate [ Cu ⁺ ] which is needed to begin precipitate.

we have , K sp = [ Cu ⁺ ] ² [ S ^2- ] = 1.0 $\times$ 10 ^-48

$\therefore$ [ Cu ⁺ ] ² = 1.0 $\times$ 10 ^-48 / [ S ^2- ]

$\therefore$ [ Cu ⁺ ] ² =   1.0 $\times$ 10 ^-48 / 2.0 $\times$ 10 ^-15

$\therefore$ [ Cu ⁺ ] ² = 5.0 $\times$ 10 ^-34

$\therefore$ [ Cu ⁺ ] = 2.2 $\times$ 10 ^-17 M

ANSWER : Concentration of Cu (I) needed to begin precipitation is 2.2 $\times$ 10 ^-17 M .