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A solution containing 10^(-2) M MgCl(2) ...

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)`

A

5

B

6

C

9

D

10

Text Solution

Verified by Experts

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

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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)]

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To a solution of 0.01M Mg^(2+) and 0.8M NH_(4)CI , and equal volume of NH_(3) is added which just gives precipitates. Calculate [NH_(3)] in solution. K_(sp) of Mg(OH)_(2) = 1.4 xx 10^(-11) and K_(b) of NH_(4)OH = 1.8 xx 10^(-5) .

Assertion (A): A solution contains 0.1M each of pB^(2+), Zn^(2+),Ni^(2+) , ions. If H_(2)S is passed into this solution at 25^(@)C . Pb^(2+), Ni^(2+), Zn^(2+) will get precpitated simultanously. Reason (R): Pb^(2+) and Zn^(2+) will get precipitated if the solution contains 0.1M HCI . [K_(1) H_(2)S = 10^(-7), K_(2)H_(2)S = 10^(-14), K_(sp) PbS =3xx 10^(-29) K_(sp) NiS = 3 xx 10^(-19). K_(sp) ZnS = 10^(-25)]

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) ,

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