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. \(1.0 \times 10^{-8} M\)
Given that the ionic product of is 2 × .
The solubility of in 0.1 M NaOH is ;
1. | 2 × M
|
2. | 1 × M
|
3. | 1 × M
|
4. | 2 × M |
The solubility product for a salt of type AB is . The molarity of its standard solution will be:
1.
2.
3.
4.
The molar solubility of in 0.1 M solution of NaF will be:
1. | 2. | ||
3. | 4. |