A solution containing `10^(-2)` M `MgCl_(2)` will just form precipitate when pH in the solution is :
`K_(sp)(Mg(OH)_(2))=10^(-12)`

Calculate pH of saturated solution Mg(OH)_(2),K_(sp) for Mg(OH)_(2) is 8.9 xx 10^(-12) .

Calculate pH of saturated solution Mg(OH)_(2),K_(sp) for Mg(OH)_(2) is 8.9 xx 10^(-12) .

Find the PH of saturated solution of Mg(OH)_2 [K_p of Mg(OH_2)= 8.9xx 10^(-12)]

The maximum pH of a solution which is having 0.10M in Mg^(2+) and from which Mg(OH)_(2) is not precipated is: (Given K_(sp)Mg(OH)_(2)=4xx10^(-11)M^(3)){log2=0.30}

The maximum pH of a solution which is 0.10 M is Mg^(2+) from which Mg(OH)_(2) is not precipitated is [ K_(sp) of Mg(OH)_(2) = 1.2 xx 10^(-11) M^(3) ]

Should Mg(OH)_(2) precipitate from a solution that is 0.01 M MgCl_(2) if the solution is also made 0.10 M in NH_(3)[K_((sp)[Mg(OH)_(2)])=1.8 xx10^(-11) , K_(b(NH_(4)OH))= 1.8 xx 10^(-5)]

Calculate pH of a saturated solution of Mg(OH)_(2) . (K_(SP)for Mg(OH)_(2)=8.9xx10^(-12))

The maximum pH of a solution which is having 0.10M in Mg^(2+) and from which Mg(OH)_(2) is not precipated is: (Given K_(sp)Mg)(OH)_(2)=4xx10^(11)M^(3)){log2=0.30}

K_(sp) of Mg(OH)_(2) is 1xx10^(-12).0.01 M MgCl_(2) will show precipitation is a solution of pH greater than :

An aqueous solution is 0.3 M in Al^(3+) and 3 M in Mg^(2+) . Select the only incorrect option. Given : (K_(sp))_(Al(OH)_(3))=3xx10^(-34) , (K_(sp))_(Mg(OH)_(2))=3xx10^(-12) ,