A solution containing 0-1 M Zn 2- and 0-01 M Cu 2- saturated with H2S- The S 2- concentration is 8-1- 10 -31 M- Will ZnS and CuS precipitate- K sp ZnS-3-0- 10 -23 and K sp CuS-8-0- 10 -4

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Common Ion Effect

A solution co...

Question

A solution containing $0.1 M$ ${Z n}^{2 +}$ and $0.01 M$ ${C u}^{2 +}$ saturated with $H_{2} S$ . The $S^{2 -}$ concentration is $8.1 \times 10^{- 31} M$ . Will $Z n S$ and $C u S$ precipitate?
$K_{s p} (Z n S) = 3.0 \times 10^{- 23}$ and
$K_{s p} (C u S) = 8.0 \times 10^{- 4}$

Open in App

Solution

For the precipitation of $M S$ the product of concentration of $M^{2 +}$ and $S^{2 -}$ will be greater than their solubility products.
Now, $[Z n^{2 +}] [S^{2 -}] = 0.1 \times 10^{- 31} \times 8.1 = 0.81 \times 10^{- 31} < K_{s p}$ of $Z n S$ and $[C u^{2 +}] [S^{2 -}] = 0.01 \times 8.1 \times 10^{- 31} = 8.1 \times 10^{- 33} < K_{s p}$ of $C u S$ .
Therefore $Z n S$ and $C u S$ will not precipitate.

Suggest Corrections

Similar questions

Q. A solution, containing

0.01 M Z n^{+ 2}

and

0.01 M C u^{2 +}

is saturated by passing

H_{2} S

gas. The

S^{- 2}

concentration is

8.1 \times 10^{- 21} M

K s p

for

Z n S

and

C u s

are

3.0 \times 10^{- 22}

and

8.0 \times 10^{- 36}

respectively. Which of the following will occur in the solution :-

Q. What

(H_{3} O^{+})

must be maintained in a saturated

H_{2} S

solution to precipitate

{P b}^{2 +}

, but not

{Z n}^{2 +}

from a solution in which each ion is present at a concentration of

0.01 M

?
(

K_{s p} H_{2} S = 1.1 \times 10^{- 22}

and

K_{s p} Z n S = 1.0 \times 10^{- 21}

)

Q. A saturated solution of

H_{2} S

0.1

M

H C l

25^{o} C

contains

S^{2 -}

ion concentration of

10^{- 23}

m o l

L^{- 1}

. The solubility product of some sulphides are

C u S = 10^{- 44}

F e S = 10^{- 14}

M n S = 10^{- 15}

C d S = 10^{- 25}

. If

0.01 M

solution of these salts in

1 M

H C l

are saturated with

H_{2} S

, which of these will be precipitated?

Q. A solution contains 0.01 M

Z n^{2 +}

and 0.01 M

C u^{2 +}

ions. It is saturated by passing

H_{2} S

gas in the solution. The

S^{2 -}

ion concentration is

9.2 \times 10^{- 22}

M . The solubility products of ZnS and CuS are

3 \times 10^{- 22}

and

8 \times 10^{- 36}

respectively.Which of the following is true ?

The minimum pH required to prevent the precipitation of ZnS in a solution containing 0.01 M $Z n C l_{2}$ and saturated with 0.10 $M H_{2} S$ is

(Given $K s p (Z n S) = 10^{- 21}, K a_{1} \times K a_{2} o f H_{2} S = 10^{- 20}$ )

Join BYJU'S Learning Program

A solution containing 0.1 M Zn2+ and 0.01 M Cu2+ saturated with H2S. The S2− concentration is 8.1×10−31 M. Will ZnS and CuS precipitate?Ksp(ZnS)=3.0×10−23 andKsp(CuS)=8.0×10−4

For the precipitation of MS the product of concentration of M2+ and S2− will be greater than their solubility products.Now, [Zn2+][S2−]=0.1×10−31×8.1=0.81×10−31<Ksp of ZnS and [Cu2+][S2−]=0.01×8.1×10−31=8.1×10−33<Ksp of CuS. Therefore ZnS and CuS will not precipitate.

A solution containing $0.1 M$ ${Z n}^{2 +}$ and $0.01 M$ ${C u}^{2 +}$ saturated with $H_{2} S$ . The $S^{2 -}$ concentration is $8.1 \times 10^{- 31} M$ . Will $Z n S$ and $C u S$ precipitate?
$K_{s p} (Z n S) = 3.0 \times 10^{- 23}$ and
$K_{s p} (C u S) = 8.0 \times 10^{- 4}$