Equilibrium MCQs for NEET — Chemistry Questions with Answers

(d) Aspirine a weak acid is unionised in acid medium due to common ion effect and completely ionised in alkaline medium.

(d) CH₃COOH (weak acid) and CH₃COONa (conjugated salt) form acidic buffer and for acidic buffer,

         pH =pK_a + log$\frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]}$ and [H⁺] = -antilog pH

          pH = -logK_a +  log$\frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]}$     [^..^. pK_a = -log K_a]

               = -log(1.8 x 10-5) + log$\frac{(0.20)}{(0.10)}$

               = 4.74 + log2

               = 4.74+0.3010 =5.041

   Now,  [H⁺] = antilog(-5.045)

                    = 9.0x10^-6 mol/L

(b) e.g. , NH₄Cl ; hydrolysis of ${\mathrm{NH}}_4^{+}$ ions, i.e., cation occurs.

(b) Quotient Q=[C]³/[A][B] = (4³ x 3 x 3)/(3³ x 1 x2) = 10.66

      [(C) = 4/3 ; (A) =1/3 ; (B) =2/3]

     Since, K_c =10, Thus, Q must decrease to attain value of Kc and therefore, [C] must decrease or [A] or [B] must increase, i.e, backward direction.

(c) K₁ = $\frac{\left[{\mathrm{XeOF}}_4\right][\mathrm{HF}{]}^2}{\left[{\mathrm{XeF}}_6\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}$                                                                   ...(i)

     K₂ = $\frac{\left[{\mathrm{XeOF}}_4\right]\left[{\mathrm{XeO}}_3{\mathrm{F}}_2\right]}{\left[{\mathrm{XeO}}_4\right]\left[{\mathrm{XeF}}_6\right]}$                                                             ....(ii)

  By Eq. (ii)/(i) we have K₂/K₁ = $\frac{\left[{\mathrm{XeO}}_3{\mathrm{F}}_2\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}{\left[{\mathrm{XeO}}_4\right][\mathrm{HF}{]}^2}$ = K_c

(b) Higher is pH, lesser is acidic nature. Also NH₄Cl (aq) is acidic and NaCN(aq) is alkaline. NaCl(aq) is neutral.

(b) Find solubility for each separately by s² = K_sp for MnS and ZnS,

108 S⁵ = K_sp for Bi₂S₃ and

4s³ = K_sp for Ag₂S.

(c) NH₃ is Lewis base.

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

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

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

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