The fertilizer which makes the soil acidic is:

1. ${\left(N{H}_4\right)}_{2}S{O}_4$

2.  Super phosphate of lime .

3. $C{H}_{3}COONa$

4. $Ca{\left(N{O}_3\right)}_2$

What is the correct relationship between the pHs of isomolar solutions of sodium oxide (pH₁), sodium sulphide (pH₂), sodium selenide (pH₃), and sodium telluride (pH₄)?

1. ${\mathrm{pH}}_1<{\mathrm{pH}}_2<{\mathrm{pH}}_3<{\mathrm{pH}}_4$

2. ${\mathrm{pH}}_1>{\mathrm{pH}}_2>{\mathrm{pH}}_3>{\mathrm{pH}}_4$

3. ${\mathrm{pH}}_1<{\mathrm{pH}}_2<{\mathrm{pH}}_3\approx {\mathrm{pH}}_4$

4. ${\mathrm{pH}}_1>{\mathrm{pH}}_2\approx {\mathrm{pH}}_3>{\mathrm{pH}}_4$

H₂S gas when passed through a solution of cations containing HCl precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is because:
 

Among the following examples, the species that behave(s) as a Lewis acid is/are: 
$B{F}_3,SnC{l}_2,SnC{l}_4$ 
1. Stannous chloride, stannic chloride
2. $B{F}_3$, stannous chloride
3. Only $B{F}_3$
4. $B{F}_3$, stannous chloride , stannic chloride

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

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. $5\times {10}^{-13}$