The molar solubility of `Al(OH)_(3)` in 0.2 M NaOH solution is `x xx 10^(-22)"mol/L"`. Given that, solubility product of `Al(OH)_(3)=2.4xx10^(-24)`. What is numerical value of x?

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)

molar solubility of Cd(OH)_(2)(Ksp=2.5xx10^(-14)) in 0.1 M KOH solution is :-

molar solubility of Cd(OH)_(2)(Ksp=2.5xx10^(-14)) in 0.1 M KOH solution is :-

The molar solubility of Zn(OH)_2 in 0.1 M NaOH solution is x xx 10^(-18)M. The value of x is _________. (Nearest integer) (Given: The solubility product of Zn(OH)_2" is " 2 xx 10^(-20) )

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

the solubility product of Al(OH)_3 is 1xx10^-36 . Calculate the solubility of Al(OH)_3.

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.