Home
Class 11
CHEMISTRY
If K(sp)" of Mg(OH)"2" is "1.2 xx 10^(-1...

If `K_(sp)" of Mg(OH)"_2" is "1.2 xx 10^(-11),` then the highest pH of the 0.1 M solution of `Mg^(2+)` ion from which `Mg(OH)_2` is not precipitated is

A

4.96

B

6.96

C

7.54

D

9.04

Text Solution

Verified by Experts

The correct Answer is:
D
`As K_(sp)" of "Mg(OH)_2=[Mg^(2+)] [OH^(-)]^2`
`[OH^(-)]^2=(1.2 xx 10^(-11))/(0.1)=1.2 xx 10^(-10) rArr [OH^(-)]=1.1 xx 10^(-5)`
`[H^+]=10^(-14)//(1.1 xx 10^(-5))=0.91 xx 10^(-9)`
`pH=-log [H^+]=-log (0.91 xx 10^(-9))=9.04`
Promotional Banner

Similar Questions

Explore conceptually related problems

What is the minimum pH of a solution of 0.1 M in Mg^(2+) from which Mg(OH)_(2) will not precipitate K_(sp) =1.2 xx 10^(-11) M^(3).

The K_(sp) of Mg(OH)_(2) is 8.9 xx 10^(-12) at 25^(2)C . the pH of solution is adjusted to 9 . How much Mg^(2+) ion will be precipitated as Mg(OH)_(2) from a 0.1M MgCl_(2) solution at 25^(@)C ? Assume that MgCl_(2) is completely dissociated.

The solubility product of Mg(OH)_(2) is 1.0 xx 10^(-12) . Concentrated aqueous NaOH solution is added to a 0.01 M aqueous solution of MgCl_(2) . The pH at which precipitation occur is -

The solubility product of Mg(OH)_2 is 1.2 xx 10^(-11) . Calculate its solubility in 0.1 M NaOH solution.

The K_(sp) of Mg(OH)_(2) is 1xx10^(-12). 0.01M Mg^(2+) will precipitate tate at the limiting pH of

K_(sp) of Mg(OH)_2 is 4.0 xx 10^(-12) . The number of moles of moles of Mg^(2+) ions in one litre of its saturated solution in 0.1 M NaOH is

K_(sp) of Mg(OH)_(2) is 1.8 xx 10^(-11) at 30^(@)C . Its molar solubility is ….......... at pH =5