Home
Class 11
CHEMISTRY
The molar solubility of PbI(2) in 0.2 M ...

The molar solubility of `PbI_(2)` in `0.2 M Pb(NO_(3))_(2)` solution in terms of solubility product, `K_(sp)`

A

`(K_(sp)//0.2)^(1//2)`

B

`(K_(sp)//0.8)^(1//3)`

C

`(K_(sp)//0.4)^(1//2)`

D

`(K_(sp)//0.8)^(1//2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`K_(SP)=[Pb^(+2)][I^(-)]^(2)`
`=(0.2+s)(2s)^(2)`
Promotional Banner

Topper's Solved these Questions

  • ACIDS & BASES

    NARAYNA|Exercise EXERCISE-I (H.W) (IONISATION OF ACIDS & BASES, DEGREE OF IONISATION & Kw)|7 Videos

  • ACIDS & BASES

    NARAYNA|Exercise EXERCISE-I (H.W) (SALT HYDROLYSIS & BUFFERS)|23 Videos

  • ACIDS & BASES

    NARAYNA|Exercise EXERCISE-I (C.W) ( BUFFER SOLUTION )|11 Videos

  • 14TH GROUP ELEMENTS

    NARAYNA|Exercise Subjective Questions|6 Videos

  • ALKANES

    NARAYNA|Exercise Integer Answer Type|7 Videos

Similar Questions

Explore conceptually related problems

The molar solubility of M(OH)_(3) in 0.4 M M(NO_(3))_(3) solution inters of solubility product of M(OH)_(3)

What is the molar solubility of Al(OH)_(3) in 0.2 M NaOH solution? Given that, solubility product of Al(OH)_(3) = 2.4 xx 10^(-2)

The solubility of CaF_(2) in a solution of 0.1M Ca(NO_(3))_(2) is

The solubility of PbI_2 in 0.1M Pb(NO_3)_2 Solution is Xxx 10^(_2) then value of X is [Given K_(sp) (PbI_2) = 8xx10^(-9)

What is the molar solubility of MgF_(2) in a 0.2 M solution of KF? (K_(sp)=8xx10^(-8))

Calculate the molar solubility of Ni(OH)_2 in 0.10 M NaOH. The solubility product of Ni(OH)_2 is 2.0xx10^(-15) .