Chemistry MCQs for NEET — Practice Questions with Answers

(c) NH₃ is Lewis base.

(b) pH of salts of weak acid and weak base is derived by the relation:

              [H⁺]=$\sqrt{K_H}$=$\sqrt{\frac{K_w}{K_a.K_b}}$

$$\begin{gathered} \mathit{\left(}\mathit{c}\mathit{\right)} \mathrm{In} \mathrm{equation}, \mathrm{X} \stackrel{ }{\rightleftharpoons } \mathrm{Y} + \mathrm{Z} \\ \mathrm{Initial} \mathrm{moles} 1 0 0 \\ \mathrm{At} \mathrm{equil}. (1-\mathrm{\alpha }) \mathrm{\alpha } \mathrm{\alpha } \\ \mathrm{where}, \mathrm{\alpha }=\mathrm{degree} \mathrm{of} \mathrm{dissociation} \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{moles} \\ =1-\mathrm{\alpha }+\mathrm{\alpha }+\mathrm{\alpha } = (1+\mathrm{\alpha }) \\ {\mathrm{P}}_{\mathrm{x}}=\left(\frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}\right){\mathrm{P}}_1 \\ {\mathrm{P}}_{\mathrm{y}}=\left(\frac{\mathrm{\alpha }}{1+\mathrm{\alpha }}\right){\mathrm{P}}_1 \\ {\mathrm{P}}_{\mathrm{z}}=\frac{\mathrm{\alpha }}{(1-\mathrm{\alpha })}{\mathrm{P}}_1 \\ {\mathrm{K}}_{{\mathrm{P}}_1}=\frac{\left[{\mathrm{P}}_{\mathrm{y}}\right]\left[{\mathrm{P}}_{\mathrm{z}}\right]}{\left[{\mathrm{P}}_{\mathrm{x}}\right]}=\frac{\left({\displaystyle \frac{\mathrm{\alpha }}{1+\mathrm{\alpha }}}\right){\mathrm{P}}_1 \times \left({\displaystyle \frac{\mathrm{\alpha }}{1+\mathrm{\alpha }}}\right){\mathrm{P}}_1}{\left({\displaystyle \frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}}\right){\mathrm{P}}_1} \\ =\frac{{\left({\displaystyle \frac{\mathrm{\alpha }}{1+\mathrm{\alpha }}}\right)}^2{\mathrm{P}}_1}{\left({\displaystyle \frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}}\right)} ............\left(\mathrm{i}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{For} \mathrm{equation}, \mathrm{A} \stackrel{ }{\rightleftharpoons } 2\mathrm{B} \\ \mathrm{Initial} \mathrm{moles} 1 0 \\ \mathrm{At} \mathrm{equil}. (1-\mathrm{\alpha }) 2\mathrm{\alpha } \\ \mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{moles} \mathrm{at} \mathrm{equilibrium} = (1+\mathrm{\alpha }) \\ {\mathrm{P}}_{\mathrm{B} }=\left(\frac{2\mathrm{\alpha }}{1+\mathrm{\alpha }}\right) {\mathrm{P}}_2 \\ {\mathrm{P}}_{\mathrm{A} }=\left(\frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}\right) {\mathrm{P}}_2 \\ {\mathrm{K}}_{{\mathrm{P}}_2}=\frac{{\left[{\mathrm{P}}_{\mathrm{B}}\right]}^2}{\left[{\mathrm{P}}_{\mathrm{A}}\right]}=\frac{\left[\left({\displaystyle \frac{2\mathrm{\alpha }}{1+\mathrm{\alpha }}}\right)\right]{\mathrm{P}}_2}{\left({\displaystyle \frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}}\right)} ...\left(\mathrm{ii}\right) \\ \mathrm{Eq}. \left(\mathrm{i}\right) \mathrm{divide} \mathrm{by} \mathrm{Eq}. \left(\mathrm{ii}\right) \\ \frac{{\mathrm{K}}_{{\mathrm{P}}_1}}{{\mathrm{K}}_{{\mathrm{P}}_2}} = \frac{{\mathrm{\alpha }}^2\times {\mathrm{P}}_1}{4{\mathrm{\alpha }}^2\times {\mathrm{P}}_2} \\ \frac{9}{1}=\frac{{\mathrm{P}}_1}{4{\mathrm{P}}_2} \\ \frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}=\frac{36}{1}=36:1 \end{gathered}$$

(a) A^pB^q $\rightleftharpoons$pA⁺ + qB^- ; Let solubility be s mol/litre, Thus

                  p.s       q.s

                    K_sp = [A⁺]^p[B^-]^q = (ps)^p.(qs)^q 

                          = p^pq^q(s)^p+q

(b) 1 mole of N₂ reacts with 3 moles of H₂ thus, for 

    N₂ + 3H₂ $\rightleftharpoons$ 2NH₃ ; (a-x) > (a-3x)

    a         a

(a-x)    (a-3x)     2x

(d) Acid $\stackrel{-{\mathrm{H}}^{+}}{\to }$conjugate base;

     Base$\stackrel{+{\mathrm{H}}^{+}}{\to }$conjugate acid.

(b)                NH₄HS(s)     $\rightleftharpoons$NH₃ (g) + H₂S(g)

            Pressure at equlibrium                  P             P

             .^.. Total pressure at equilibrium = 2P =1.12 atm

                                                       P =1.12/2 atm

                ^..^. K_p = $P{\mathit{\text{'}}}_{N{H}_{\mathit{3}}}\mathit{ }\mathrm{x}\mathit{ }P{\mathit{\text{'}}}_{H_{\mathit{2}}S}$ 

                         K_p = (1.12/2) x (1.12/2) = 0.3136 atm²

(c) Hg exists as ${\mathrm{Hg}}_2^{2+}$ and not as Hg⁺. Thus ${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Hg}}_2^{2+}\right][{\mathrm{I}}^{-}{]}^2$

(b) In aqueous solution of 10^-8 M HCl, [H⁺] is based upon the concentration of H+ ion of 10^-8 M HCl and concentration of H+ ion of water.

             K_w of H₂O = 10^-14 = [H⁺][OH^-]

       or [H⁺] = 10^-7 M (due to its neutral behaviour)

  So , in aqueous solution of 10^-8 M HCl,

          [H⁺] = [H⁺] of HCl + [H⁺] of water = 10^-8 + 10^-7 = 11x10^-8 M$\approx$1.10 x 10^-7

(a) Lower is the value of K, lesser will be the tendency to show forward reaction.