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 solubility of $\mathrm{Ni}{\left(\mathrm{OH}\right)}_2$ in 0.1 M NaOH is:

K_sp ($\mathrm{Ni}{\left(\mathrm{OH}\right)}_2$)= 2x ${10}^{-15}$

1. 2 x ${10}^{-8}$ M.

2. 1 x ${10}^{-13}$ M

3. 1 x ${10}^8$ M 

4. 2 x ${10}^{-13}$ 

At room temperature, MY and NY₃, two nearly insoluble salts, have the same K_sp values of 6.2 × 10^-13. The true statement regarding MY and NY₃ is:

If the solubility of a ${\mathrm{M}}_2\mathrm{S}$ salt is $3.5$ $\times$ ${10}^{-6}$ mol litre^-1 , then the solubility product of ${\mathrm{M}}_2\mathrm{S}$ will be:

1. $1.7$ $\times$ ${10}^{-6}$ mol³ litre^-3

2. $1.7$ $\times$ ${10}^{-16}$ mol³ litre^-3

3. $1.7$ $\times$ ${10}^{-18}$ mol³ litre^-3

4. $1.7$ $\times$ ${10}^{-12}$ mol³ litre^-3

The minimum volume of water required to dissolve 1g of calcium sulphate at 298 K is

(For CaSO₄ , K_sp is 9.1 × 10^–6)

1. 1.22 L

2. 0.69 L

3. 2.44 L

4. 1.87 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:

The solubility product of mercurous iodide is $4.5\times {10}^{-29}$. The solubility of mercurous iodide will be:

1. $6.5\times {10}^{-7}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

2. $4.09\times {10}^{-8}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

3. $4.09\times {10}^{-7}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

4. $6.5\times {10}^{-8}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

The solubility of BaSO₄ in water is $2.42$ $\times$ ${10}^{-3}$ g/ litre at 298 K. The value of the solubility product will be: (Molar mass of BaSO₄ = 233 gmol^–1)

A saturated solution of Ba(OH)₂ has a pH of 12. The value of its K_sp will be:
1. $4.00\times {10}^{-6}$ $M^3$
2. $4.00\times {10}^{-7}$ $M^3$
3. $5.00\times {10}^{-6}$ $M^3$
4. $5.00\times {10}^{-7}$ $M^3$

The maximum concentration of equimolar solutions, of ferrous sulphate and sodium sulphide, so that when mixed in equal volumes, there is no precipitation of iron sulphide, will be:

(For iron sulphide, K_sp = 6.3 × 10^–18). 

$1.$ $5.02$ $\times {10}^{-9}$ $\mathrm{M}$

$2.$ $5.02$ $\times$ ${10}^9$ $\mathrm{M}$

$3.$ $2.$ $25$ $\times$ ${10}^{-13}$ $\mathrm{M}$

$4.$ $\mathrm{Can}\text{'}\mathrm{t}$ $\mathrm{predict}$