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

Byju's Answer

Standard XII

Chemistry

Heat of Formation

The minimum p...

Question

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

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

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

Open in App

Solution

Step 1:
To prevent ionic product of ions should not exceed $K_{s p}$ .
So, $[S^{2 -}] = \frac{K_{s p}}{[{Z n}^{2 +}]} = \frac{10^{- 21}}{10^{- 2}} = 10^{- 19}$
Step 2:
$H_{2} S ⟶ 2 H^{+} + S^{2 -}$
$K_{a} = \frac{{[H^{+}]}^{2} [S^{2 -}]}{[H_{2} S]}$
${[H^{+}]}^{2} = 0.01 a s [H^{+}] = 0.1 M$
So, pH = 1
Hence, option (D) is correct.

Suggest Corrections

Similar questions

Q. What is the minimum

p H

required to prevent the precipitation of

Z n S

in a solution that is

0.01

M Z n C l_{2}

and saturated with

0.10 M H_{2} S

?
[Given :

K_{s p} = 10^{- 21}, K_{a_{1}} \times K_{a_{2}} = 10^{- 20}]

Q. What is the minimum

p H

required to prevent the precipitation of

Z n S

in a solution that is

0.01 M Z n C l_{2}

and saturated with

0.10 M H_{2} S

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

Q. An acidified solution of

0.05 M Z n^{2 +}

is saturated with

0.1 M H_{2} S

. What is the minimum molar concentration

(M)

H^{+}

required to prevent the precipitation of

Z n S

?
Use

K_{S P} (Z n S) = 1.25 \times 10^{- 22}

and overall dissociation constant of

H_{2} S

K_{N E T} = K_{1} K_{2} = 1 \times 10^{- 21}

Q. 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}

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

The minimum pH required to prevent the precipitation of ZnS in a solution containing 0.01 M ZnCl2 and saturated with 0.10 MH2S is (Given Ksp(ZnS)=10−21,Ka1×Ka2ofH2S=10−20)

Step 1: To prevent ionic product of ions should not exceed Ksp. So, [S2−]=Ksp[Zn2+]=10−2110−2=10−19 Step 2: H2S⟶2H++S2− Ka=[H+]2[S2−][H2S] [H+]2=0.01as[H+]=0.1M So, pH = 1Hence, option (D) is correct.

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