The solubility product of \(BaSO_4\) in water is \(1.5 \times 10^{-9} \). The molar solubility of \(BaSO_4\) in 0.1 M solution of Ba(NO3)2 in- 
1. \(2.0 \times 10^{-8} M\)
2. \(0.5 \times 10^{-8} M\)
3. \(1.5 \times 10^{-8} M\)
4. \(1.0 \times 10^{-8} M\)

Subtopic:  Solubility Product |
 80%
From NCERT
NEET - 2022
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Given that the ionic product of NiOH2 is 2 x 10-15 .
The solubility of NiOH2 in 0.1 M NaOH is ;

1. 2 x 10-8 M

2. 1 x 10-13 M

3. 1 x 108

4. 2 x 10-13 M

Subtopic:  Solubility Product |
 68%
From NCERT
NEET - 2020
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The solubility product for a salt of type AB is 4×10-8.  The molarity of its standard solution will be:
1. 2×10-4 mol/L

2. 16×10-16 mol/L

3. 2×10-16 mol/L

4. 4×10-4 mol/L

Subtopic:  Solubility Product |
 80%
From NCERT
NEET - 2020
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The molar solubility of CaF2 (Ksp=5.3 x 10-11) in 0.1 M solution of NaF will be:

1. 5.3 x 10-11 mol L-1

2. 5.3 x 10-8 mol L-1

3. 5.3 x 10-9 mol L-1

4. 5.3 x 10-10 mol L-1

Subtopic:  Solubility Product |
 65%
From NCERT
NEET - 2019
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