$150$ mL of $0.0008 M$ ammonium sulphate is mixed with $50$ mL of $0.04 M$ calcium nitrate. The ionic product of $C a S O_{4}$ will be:
$(K_{s P} = 2.4 \times 10^{- 5}$ for $C a S O_{4})$

< K_{s p}

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> K_{s p}

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\approx K_{s p}

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None of these

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Solution

The correct option is B $< K_{s p}$
$[C a^{2 +}] = 0.04 M \times \frac{50 m L}{150 m L + 50 m L} = 0.01 M$

$[S O_{4}^{2 -}] = 0.0008 M \times \frac{150 m L}{150 m L + 50 m L} = 0.0006 M$

The ionic product of $C a S O_{4}$ ,
$Q = [C a^{2 +}] [S O_{4}^{2 -}]$

$Q = 0.01 M \times 0.0006 M$

$Q = 6 \times 10^{- 6}$

The solubility product $K_{s p} = 2.4 \times 10^{- 5}$

Hence, $Q < K_{s p}$ and no precipitate will be formed.

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150 mL of 0.0008 M ammonium sulphate is mixed with 50 mL of 0.04 M calcium nitrate. The ionic product of CaSO4 will be:(KsP=2.4×10−5 for CaSO4)

The correct option is B <Ksp[Ca2+]=0.04 M×50 mL150 mL+50 mL=0.01 M[SO2−4]=0.0008 M×150 mL150 mL+50 mL=0.0006 MThe ionic product of CaSO4 , Q=[Ca2+][SO2−4]Q=0.01 M×0.0006 MQ=6×10−6The solubility product Ksp=2.4×10−5Hence, Q<Ksp and no precipitate will be formed.

$150$ mL of $0.0008 M$ ammonium sulphate is mixed with $50$ mL of $0.04 M$ calcium nitrate. The ionic product of $C a S O_{4}$ will be:
$(K_{s P} = 2.4 \times 10^{- 5}$ for $C a S O_{4})$