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 $\mathrm{Ni}{\left(\mathrm{OH}\right)}_2$ is 2 × ${10}^{-\mathrm{15<mspace\; width="1em"></mspace>}}$. The solubility of $\mathrm{Ni}{\left(\mathrm{OH}\right)}_2$ in 0.1 M NaOH is ;
1. 2 × ${10}^{-8}$ M

What is the molarity of the standard solution if the solubility product for a salt of type AB is $4\times {10}^{-8}$?

1. $2\times {10}^{-4}$ $\mathrm{mol}/\mathrm{L}$

2. $16\times {10}^{-16}$ $\mathrm{mol}/\mathrm{L}$

3. $2\times {10}^{-16}$ $\mathrm{mol}/\mathrm{L}$

4. $4\times {10}^{-4}$ $\mathrm{mol}/\mathrm{L}$

The molar solubility of ${\mathrm{CaF}}_2$ $({\mathrm{K}}_{\mathrm{sp}}=5.3$ $\mathrm{\times }$ ${10}^{-11})$ in 0.1 M solution of NaF will be: