In the reaction A(g) + 2B(g) ⇌ 2C(g) + D(g), the initial concentration of B is twice that of A and, at equilibrium, the concentrations of A and D are equal. The value of the equilibrium constant will be:

1.	4	2.	16
3.	2	4.	1

$2\mathrm{HI}\stackrel{}{\rightleftharpoons }{\mathrm{H}}_2+{\mathrm{I}}_2$ for 22% dissociation what would be the value of equilibrium constant?

1. 0.04

2. 0.0198

3. 0.5

4. 1.2527

The degree of hydrolysis of the salt is independent of the concentration of a solution: 
1. NH₄CI
2. NH₄CN
3. (NH₄)₂SO₄
4. All of the above

The dissociation of solid ${\mathrm{NH}}_4\mathrm{SH}$ in a closed container produces a pressure of 4 atm at $95°\mathrm{C}$, then ${\mathrm{K}}_{\mathrm{p}}$ for the reaction will be:

${\mathrm{NH}}_4\mathrm{SH}\left(\mathrm{s}\right)\stackrel{}{\rightleftharpoons }{\mathrm{NH}}_3\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)$

The number of ${\mathrm{H}}^{\oplus }$ ions present in 1 ml of a solution whose pH= 4 , is given as:

($N_A=$ $Avogadro\text{'}s$ $number$)

1. ${10}^{-7}{N}_A$

2. ${10}^{-8}{N}_A$

3. ${10}^{-16}{N}_A$

4. ${10}^{-14}{N}_A$

If the pH of an acidic buffer is 5.7 and ${\mathrm{pK}}_{\mathrm{a}}$ is 5 then the ratio of $\frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]}$ will be:

On adding NH₄CI to an NH₄OH solution, the pH of the solution will

Consider the following two reactions : 

$\mathrm{NO}\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\stackrel{}{\rightleftharpoons }{\mathrm{NO}}_2\left(\mathrm{g}\right);$ ${\mathrm{K}}_1=$ $4\times {10}^{-3}$

$2{\mathrm{NO}}_2\left(\mathrm{g}\right)\stackrel{}{\rightleftharpoons }2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right);$ ${\mathrm{K}}_2=?$

If ${\mathrm{K}}_1$ $\mathrm{and}$ ${\mathrm{K}}_2$ are equilibrium constants, the value of ${\mathrm{K}}_2$ will be:

The concentration of $\left[{\mathrm{H}}^{+}\right]$ ion in a solution containing 0.1 M HCN and 0.2 M NaCN will be:

(${\mathrm{K}}_{\mathrm{a}}$ for HCN = $6.2\times {10}^{-10}$)

1. 3.1$\times {10}^{10}$

2. $6.2\times {10}^5$

3.$$ $6.2\times {10}^{-10}$

4. $3.1\times {10}^{-10}$

If the molar concentration of ${\mathrm{SnCl}}_4$ is $1.5$ $\times$ ${10}^{-3}$ mol L^–1, the concentration of chloride ions will be: