Home
Class 12
CHEMISTRY
Solubility of Cd(OH)(2) in pure water is...

Solubility of `Cd(OH)_(2)` in pure water is `1.84xx10^(-5)"mole"//L` Calculate its solubility in a buffer solution of `pH=12`.

A

`1.84 xx 10^(-9) M`

B

`(2.49)/(1.84) xx 10^(-9) M`

C

`6.23 xx 10^(-11) M`

D

`2.49 xx 10^(-10) M`

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Topper's Solved these Questions

  • JEE MAINS

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise QUESTION|1 Videos
  • JEE MAIN

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise CHEMISTRY|146 Videos
  • JEE MAINS 2020

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise CHEMSITRY|23 Videos

Similar Questions

Explore conceptually related problems

The solubility of Pb(OH)_(2) in water is 6.7xx10^(-6) M. Calculate the solubility of Pb(OH)_(2) in a buffer solution of pH=8 .

The solubility of Pb(OH)_(2) in water is 6.7xx10^(-6) M. Calculate the solubility of Pb(OH)_(2) in a buffer solution of pH=8 .

The solubility of Pb(OH)_(2) in water is 6.7xx10^(-6) M. Calculate the solubility of Pb(OH)_(2) in a buffer solution of pH=8 .

The solubility of Pb(OH)_(2) in water is 6.7xx10^(-6) M. Calculate the solubility of Pb(OH)_(2) in a buffer solution of pH=8 .

The solubility product of SrF_(2) in water is 8xx10^(-10) . Calculate its solubility in 0.1M NaF aqueous solution.

The molar solubility of Cd(OH)_(2) is 1.84xx10^(-5)M in water. The expected solubility of Cd(OH)_(2) in a buffer solution of pH=10 is 2.49xx10^(-x)M . The numerical value of x is.

The molar solubility of Cd(OH)_(2) is 1.84xx10^(-5)M in water. The expected solubility of Cd(OH)_(2) in a buffer solution of pH = 12 is

The solubility product of SrF_(2) in water is 8 xx 10^(-10) Calculate its solubility in 0.1 M of aqueous NaF solution. If its solubility is expressed as y xx 10^(-8) then what is the value of 'y' ?

The solubility of Mg( OH)_2 in pure water is 9.57 xx 10 ^(-3) g L^(-1) .Calculate its solubility I (g L^(-1)) in 0.02 M Mg (NO_3)_2 solutions.

The solubility of Sb_(2)S_(3) in water is 1.0 xx 10^(-5) mol/letre at 298K. What will be its solubility product ?