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, as per 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. $\log \frac{\left[{\mathrm{In}}^{-}\right]}{\left[\mathrm{HIn}\right]}=\; pH\; +\hspace{0.17em}{\mathrm{pK}}_{\mathrm{In<mspace\; width="1em"></mspace>}}$

4. None 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 $\mathrm{Q}=\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}$.
Under what conditions will the reaction proceed from right to left?

1. Q = K_C
2. Q < K_C
3. Q > K_C
4. Q = 0

(where K_C is the equilibrium constant)