Question

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

Answer 1

Answer:

Since K

2

is much smaller than K

1

, H

+

are largely formed from first dissociation

H

2

S⇌[H

+

]+[HS

−

];K

1

K

1

=

[H

2

S]

[H

+

][HS

2

−

]

=

[H

2

S]

[H

+

]

2

[H

+

]=

K

1

[H

2

S]

=10

−4

pH=−log[H

+

]=4 BRI B OPTION IS CORRECT