The solubility of Ca(OH)₂ in water is: 

[Given: K_sp Ca(OH)₂ in water = 5.5 × 10^–6]

1. 1.77 × 10^–6

2. 1.11 × 10^–6

3. 1.11 × 10^–2

4. 2.77 × 10^–2

If solubility product of $Z{r}_3{\left(P{O}_4\right)}_4$ is denoted by $K_{Sp}$ and its molar solubility is denoted by S, then which of the following relation between S and $K_{Sp}$ is correct ?
1. $S<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>{\left(\frac{K_{sp}}{216}\right)}^{1/7}$

2. $S<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>{\left(\frac{K_{sp}}{6912}\right)}^{1/7}$

3. $S<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>{\left(\frac{K_{sp}}{144}\right)}^{1/6}$

4. $S<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>{\left(\frac{K_{sp}}{929}\right)}^{1/9}$

What is the molar solubility of aluminum hydroxide (Al(OH)₃) in a 0.2 M sodium hydroxide (NaOH) solution?
[Given: Solubility product of Al(OH)₃ is 2.4 × 10^–24]

1. 3 × 10^–22
2. 3 × 10^–19
3. 12 × 10^–21
4. 12 × 10^–22

The molar solubility of Cd(OH)₂ is 1.84 × 10^–5 M in water. The expected solubility of Cd(OH)₂ in a buffer solution of pH = 12 is:

The solubility product of silver bromide is 5.0 x 10^-13. The quantity of potassium bromide (molar mass taken as 120 g mol ⁻¹ ) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is-

1. 5.0 x 10 ⁻⁸ g

2. 1.2 x 10 ⁻¹⁰ g

3. 1.2 x 10^-9 g

4. 6.2 x 10⁻⁵ g

At 25°C, the solubility product of Mg(OH)₂ is 1.0 × 10^-11. At which pH, will Mg²⁺ ions start precipitating in the form of Mg(OH)₂ from a solution of 0.001 M Mg²⁺ ions?

In a saturated solution of the sparingly soluble electrolyte AgIO₃ (molecular mass = 283) the equilibrium is represented by–
${\mathrm{AgIO}}_3\left(\mathrm{(s)}\right)\rightleftharpoons {\mathrm{Ag}}^{+}\left(\mathrm{(aq)}\right)+<mspace\; width="1em"></mspace>{{\mathrm{IO}}_3}^{-}\left(\mathrm{(aq)}\right)$
If the solubility product constant K_sp of AgIO₃ at a given temperature is 1.0 × 10^-8, Calculate the mass of AgIO₃ contained in 100 ml of its saturated solution.

1. 28.3 × 10^–2 g

2. 2.83 × 10^–3 g

3. 1.0 × 10^–7 g

4. 1.0 × 10^–4 g

The solubility product of Pbl₂ is 8.0 × 10^–9 . The solubility of lead iodide in 0.1 molar solution of lead nitrate is x × 10^–6 mol/L. The value of x is:

(Rounded off to the nearest integer)$\text{\hspace{0.17em}[Given:\hspace{0.17em}}\sqrt{2}=1.41\text{\hspace{0.17em}]\hspace{0.17em}}$