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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
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

` 7.1 xx 10 ^(-4) M`

B

` 7. 6 xx 10 ^(-3) M`

C

` 6. 67 xx 10 ^(-3) M`

D

` 5.6 xx 10 ^(-4) M`

Text Solution

Verified by Experts

The correct Answer is:
B
` Cu(OH)_2 +underset( 1-4s ) (4NH_3) hArr[underset(S) (Cu(NH_3)_4)]^(2+) +underset( 2s) (2OH,K)`
` K= K_f K_(sp) =([OH^(-) ]^(2)[Cu(NH_3)_4]^(2+))/( [NH_3]^(4)) `
` 1.1 xx 10^(13) xx 1.6 xx 10^(-19) =( (2s)^(2)(s))/( (1)^(4))`
` rArr S= 7.6 xx10^(-3) `
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