Question

Calculate the molar solubility of Copper (II) sulfide (Ksp = 6 x 10^-37) in a
solution in which pH is held constant at:
a)2
b)9
Hydrogen sulfide (K1 = 9.6 x 10^-8 & K2 = 1.3 x 10^-14)
(Briefly discuss your results)​

Answer 1

Answer:

but the class is a great way for me and I think that there are some things and I have

Explanation:

updoyxyodoyxohcoctozyoxoydgdupff was the first to be said in a statement that the government was not authorized for

Answer 2

Given:

$K_{sp} = 6 \times 10^{-37}$

To find:

Molar solubility of $CuS$

Explanation:

The reaction is give as:

$CuS \leftrightharpoons Cu^2^+ + S^2^-$

Solubility product equilibrium constant expression is given by:

$K_sp = [Cu^2^+][S^2^-]$

Let the equilibrium concentration of ions be x M

$6 \times 10^{-37} = x^{2}$

$\sqrt{6\times 10^{-37}} = x$ $= 7.74 \times 10^{-18}$ M

Therefore molar solubility of CuS $= 7.74 \times 10^{-18}$ M

a) At pH=2 ; which is acidic , the solubility of salt increases due to Le Chateleir principle.

b) At pH=9 ; which is basic, the solubility of salt decreases due to addition  

of $OH^-$ions and Le Châtelier's Principle says that the position of equilibrium will move to the left.