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For a general reaction given below, the ...

For a general reaction given below, the value of solubility product can be given us
`{:(A_(x)B_(y),=xA^(+y),+yB^(-x)),(a,0,0),(a-s,xs,ys):}`
`K_(sp)=(xs)^(x).(ys)^(y) (or) K_(sp)=x^(x)y^(y) (S)^(x+y)`
Solubility product gives us not only an idea about the solubility of an electrolyte in a solvent but also helps in explaining concept of precipitation and calculation `[H^(+)]` ion, `[OH^(-)`] ion. It is also useful in qualitative analysis for the idetification and separation of basic radicals
Potussium chromate is slowly aded toa solution containing 0.20M Ag `NO_(3)`, and 0.20M `Ba(NO_(3))_(2)`. Describe what happensif the `K_(sp)` for `Ag_(2),CrO_(4)`, is `1.1 xx 10^(-12)` and the `K_(sp)` of `BaCiO_(4)`, is `1.2 xx 10^(-10)`,

A

The ` Ag_2CrO_4` pecipitates first out of solution and then `BaCrO_4 ` preciptates.

B

The `BaCrO_4` pecipitates first out of solution and then ` Ag_2CrO_4` preciptates

C

Both `Ag_2CrO_4 and BaCrO_4` precipitate simultaneously out of solution

D

Neither `Ag_2CrO_4 " nor " BaCrO_4` precipitates

Text Solution

Verified by Experts

The correct Answer is:
A
` [CrO_4^(2-)]_(Ag^(+)) =(Ksp)/( [Ag^(+) ]^(2)) =(1.1xx 10^(-12))/(0.2) =5.5 xx 10 ^(-12) `
` [CrO_4^(2-) ]_(Ba^(+2) ) =(Ksp)/([Ba^(+2)]) =(1.2xx 10^(-10))/( 0.2 0 ) = 6 xx 10^(-10) `
` therefore Ag_2 Cr O_4 ` ppts first
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Knowledge Check

  • For a general reaction given below, the value of solubility product can be given us {:(A_xB_y ,=xA^(+y) +,yB^(-x) ),( a,0,0),(a-x,xs,ys):} K_(sp) =(xs) ^(x) . (ys) ^(y) (or) K_(sp) =x^(x) y^(y) (S) ^(x+y) Solubility product gives us not only an idea about the solubility of an electrolyte in a solvent but also helps in explaining concept of precipitation and calculation [H^(+) ] ion , [OH^(-)] ion, It is also useful in qualitative analysis for the identification and separation of basic radicals. Potassium , chromate is slowly added to a solution containing 0.2 M AgNO_3 and 0.2 M Ba(NO_3) _2 . Describe what happens if the K_(sp) " for " Ag_2CrO_4 " is " 1.1 xx 10 ^(-12) and " the " K_(sp) " of " BaCrO_4 " is " 1.2 xx 10 ^(-10 )

    A
    The ` Ag_2CrO_4` pecipitates first out of solution and then `BaCrO_4 ` preciptates.
    B
    The `BaCrO_4` pecipitates first out of solution and then ` Ag_2CrO_4` preciptates
    C
    Both `Ag_2CrO_4 and BaCrO_4` precipitate simultaneously out of solution
    D
    Neither `Ag_2CrO_4 " nor " BaCrO_4` precipitates

  • For a general reaction given below, the value of solubility product can be given us {:(A_(x)B_(y),=xA^(+y),+yB^(-x)),(a,0,0),(a-s,xs,ys):} K_(sp)=(xs)^(x).(ys)^(y) (or) K_(sp)=x^(x)y^(y) (S)^(x+y) Solubility product gives us not only an idea about the solubility of an electrolyte in a solvent but also helps in explaining concept of precipitation and calculation [H^(+)] ion, [OH^(-) ] ion. It is also useful in qualitative analysis for the idetification and separation of basic radicals What is the molar solubility of Cu(OH)_(2) , in 1.0 M NH_(3) if the deep blue complex ion [Cu(NH_(3))_(4)]^(2+) is formed. The K_(sp), of Cu(OH)_(2) , is 1.6xx 10^(-19) and K_(3) , of [Cu(NH_(3))_(4) is 1.1 xx 10^(13)

    A
    `7.1xx10^(-4)`M
    B
    `7.1xx10^(-4)`M
    C
    `7.6xx10^(-3)`M
    D
    `5.6xx10^(-4)`M

  • For a general reaction given below, the value of solubility product can be given us {:(A_(x)B_(y),=xA^(+y),+yB^(-x)),(a,0,0),(a-s,xs,ys):} K_(sp)=(xs)^(x).(ys)^(y) (or) K_(sp)=x^(x)y^(y) (S)^(x+y) Solubility product gives us not only an idea about the solubility of an electrolyte in a solvent but also helps in explaining concept of precipitation and calculation [H^(+)] ion, [OH^(-) ] ion. It is also useful in qualitative analysis for the idetification and separation of basic radicals Which metal sulphides can be precipitated from a solution that is O.01M in Na^(+), Zn^(2+), Pb^(2+) and Cu^(2+) and 0.1OM in H_(2),S at a pH of 1.0? {:("Metal sulphide",K_(sp)),(S,3xx10^(16)),(ZnS,3xx10^(-2)),(PbS,3xx10^(-7)),(CuS,3xx10^(-16)):}

    A
    Cus
    B
    `Na_(2)S`
    C
    Zns, Obs, Cus
    D
    Pbs , Cus

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