Ksp of CdS-8- 10-27- Ksp of ZnS-1 - 10-21- Ka of H2S-1 - 10-21-When ZnS starts precipitating- what concentration of Cd2- is left-

K_sp of ZnS= 10^ 21=[Zn^2+][S^2 ]=0.1×[S^2 ] so [S^2 ]= 10^ 20 now for CdS [K_sp of CdS=8× 10^ 27=[Cd^2+][S^2 ]=[Cd^2+]×10^ 20 so [Cd^2+] left = 8× 10^ 7M ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Chemistry

Solubility Product

Ksp of CdS=8×...

Question

$K_{s p}$ of $C d S = 8 \times 10^{- 27}$ , $K_{s p}$ of $Z n S = 1 \times 10^{- 21}$ , $K_{a}$ of $H_{2} S = 1 \times 10^{- 21}$ .
When $Z n S$ starts precipitating, what concentration of $C d^{2 +}$ is left?

8 \times 10^{- 7} M

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.1 M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 \times 10^{- 10} M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 \times 10^{- 9} M

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $8 \times 10^{- 7} M$
$K_{s p}$ of $Z n S = 10^{- 21} = [Z n^{2 +}] [S^{2 -}] = 0.1 \times [S^{2 -}]$
so $[S^{2 -}] = 10^{- 20}$
now for $C d S$
[ $K_{s p}$ of $C d S = 8 \times 10^{- 27} = [C d^{2 +}] [S^{2 -}] = [C d^{2 +}] \times 10^{- 20}$
so $[C d^{2 +}]$ left = $8 \times 10^{- 7} M$

Suggest Corrections

Similar questions

K_{s p}

C d S = 8 \times 10^{- 27}

K_{s p}

Z n S = 1 \times 10^{- 21}

K_{a}

H_{2} S = 1 \times 10^{- 21}

.
When

Z n S

starts precipitating, what concentration of

C d^{2 +}

is left?

K_{s p}

C d S = 8 \times 10^{- 27}

K_{s p}

Z n S = 1 \times 10^{- 21}

K_{a}

H_{2} S = 1 \times 10^{- 21}

.
When

Z n S

starts precipitating, what concentration of

C d^{2 +}

is left?

K_{s p}

C d S = 8 \times 10^{- 27}

K_{s p}

Z n S = 1 \times 10^{- 21}

K_{a}

H_{2} S = 1 \times 10^{- 21}

.
What

[H^{+}]

must be maintained in saturated

H_{2} S (0.1 M)

to precipitate

C d S

but not

Z n S

, if

[C d^{2 +}] = [Z n^{2 +}] = 0.1 M

initially?

K_{s p}

C d S = 8 \times 10^{- 27}

K_{s p}

Z n S = 1 \times 10^{- 21}

K_{a}

H_{2} S = 1 \times 10^{- 21}

.
What

[H^{+}]

must be maintained in saturated

H_{2} S (0.1 M)

to precipitate

C d S

but not

Z n S

, if

[C d^{2 +}] = [Z n^{2 +}] = 0.1 M

initially?

Q. Maximum

p H

that must be maintained in a saturated

H_{2} S

(0.1 M) to precipitate

C d S

but not

Z n S

x

and concentration of

C d^{2 +}

ion at that

p H

y \times 10^{- 7}

then

x + y =

Given

[C d^{2 +}] = [{Z n}^{2 +}] = 0.1 M

initially

K_{s p} (C d S) = 8 \times 10^{- 27}

K_{s p} (Z n S) = 1 \times 10^{- 21}

K_{a 1} (H_{2} S) = 1.0 \times 10^{- 8}

K_{a 2} (H_{2} S) = 1.0 \times 10^{- 13}

Join BYJU'S Learning Program

Ksp of CdS=8×10−27, Ksp of ZnS=1×10−21, Ka of H2S=1×10−21. When ZnS starts precipitating, what concentration of Cd2+ is left?

The correct option is A 8×10−7 MKsp of ZnS=10−21=[Zn2+][S2−]=0.1×[S2−] so [S2−]=10−20 now for CdS [Ksp of CdS=8×10−27=[Cd2+][S2−]=[Cd2+]×10−20 so [Cd2+] left = 8×10−7M

$K_{s p}$ of $C d S = 8 \times 10^{- 27}$ , $K_{s p}$ of $Z n S = 1 \times 10^{- 21}$ , $K_{a}$ of $H_{2} S = 1 \times 10^{- 21}$ .
When $Z n S$ starts precipitating, what concentration of $C d^{2 +}$ is left?