The ratio of solubility of \(\mathrm{AgCl}\) in \(0.1~ \mathrm{M} ~\mathrm{KCl}\) solution to the solubility of \( \mathrm{Ag} \mathrm{Cl}\) in water is:
(Given : Solubility product of \(\mathrm{AgCl}=10^{-10}\) )
1. \(10^{-4}\)
2. \(10^{-6}\)
3. \(10^{-9}\)
4. \(10^{-5}\)
Subtopic:  Solubility Product |
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The solubility product of \(\mathrm{BaSO_4}\) in water is \(1.5 \times 10^{-9} \). The molar solubility of \(\mathrm{BaSO_4}\) 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 |
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Given that the ionic product of NiOH2 is 2 × 10-15 .
The solubility of NiOH2 in 0.1 M NaOH is ;

1. 2 × 10-8 M

2. 1 × 10-13 M

3. 1 × 108

4. 2 × 10-13 M
Subtopic:  Solubility Product |
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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 |
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The molar solubility of CaF2 (Ksp=5.3 × 10-11) in 0.1 M solution of NaF will be:

1. 5.3 × 10-11 mol L-1 2. 5.3 × 10-8 mol L-1
3. 5.3 × 10-9 mol L-1 4. 5.3 × 10-10 mol L-1
Subtopic:  Solubility Product |
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