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150 mL of 0.0008 M ammonium sulphate is ...

150 mL of 0.0008 M ammonium sulphate is mixed with 50 mL of 0.04 M calcium nitrate. The ionic product of `CaSO_(4)` will be : `(K_(sp)=2.4xx10^(-5) for CaSO_(4))`

A

`K_(sp)`

B

`lt K_(sp)`

C

`gt K_(sp)`

D

None

Text Solution

Verified by Experts

The correct Answer is:
B
`M^(1) = (M_(1)V_(1))/(V_(1) + V_(2))`
`[Ca^(+2)][SO_(4)^(2-)] = ((0.04 xx 50)/(150 + 50))((0.0008 xx 150)/(150 + 50)) lt ksp`
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