4 gm of NaOH is dissolved in 1000 ml of water. The ${\mathrm{H}}^{+}$ ion concentration will be:

1.	10-1 M	2.	10-13 M
3.	10-4 M	4.	10-10 M

The solubility product of a sparingly soluble salt AX₂ is 3.2 ×10^–11. Its solubility (in moles/litre) is:

1. 3.1×10^–4

2. 2 × 10^–4

3. 4 × 10^–4

4. 5.6 × 10^–6

A compound BA₂ has \(K_{sp} = 4\times 10^{-12}\)$.$Solubility of this compound will be:

The rapid change of pH near the stoichiometric point of an acid-base titration is the basis of indicator detection. pH of the solution is related to the ratio of the concentrations of the conjugate acid (HIn) and base (In–) forms of the indicator by the expression:

1. $\log \frac{\left[\mathrm{HIn}\right]}{\left[{\mathrm{In}}^{-}\right]}={\mathrm{pK}}_{\mathrm{In}}-\mathrm{pH}$

2. $\log \frac{\left[\mathrm{HIn}\right]}{\left[{\mathrm{In}}^{-}\right]}=\mathrm{pH}-{\mathrm{pK}}_{\mathrm{In}}$

3. $\log \frac{\left[{\mathrm{In}}^{-}\right]}{\left[\mathrm{HIn}\right]}=-pH\; + {\mathrm{pK}}_{\mathrm{In }}$

4. All of the above.

For a reaction , ${\mathrm{BaO}}_2\left(\mathrm{s}\right)\rightleftharpoons \mathrm{BaO}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$; ∆H = + ve. At equilibrium condition, the pressure of O₂ depends on the:

1. Increased mass of BaO₂

2. Increased mass of BaO

3. Increased temperature on equilibrium.

4. Increased mass of BaO₂ and BaO both.

If the solubility of MX₂ – type electrolytes is 0.5 × 10^–4 Mole/lit. then K_sp of electrolytes will be:

1. $5\times {10}^{-12}$

2. $25\times {10}^{-10}$

3. $1\times {10}^{-13}$

4. $5\times {10}^{-13}$

The compound with the highest pH among the following is:

1. CH₃COOK

2. Na₂CO₃

3. NH₄Cl

4. NaNO₃

If a solution of 0.1 N NH₄OH and 0.1 N NH₄Cl has pH 9.25, then pK_b of NH₄OH is:

1. 9.25

2. 4.75

3. 3.75

4. 8.25

A compound among the following that cannot be classified as a protonic acid is:

The reaction quotient (Q) for the reaction:

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

is given by $\mathrm{Q}=\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}$. The reaction will proceed from right to left if:
1. Q = K_C
2. Q < K_C
3. Q > K_C
4. Q = 0

(where K_C is the equilibrium constant)