Chemistry MCQs for NEET — Practice Questions with Answers

(d) Key Idea (i) For basic buffer,

$$\begin{gathered} \mathrm{pOH} = \mathrm{p}{K}_{\mathrm{b}} + \log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{base}\right]} \\ \left(\mathrm{ii}\right) \mathrm{pH} + \mathrm{pOH} =14 \\ \mathrm{Given}, K_{\mathrm{b}} = 1 \times {10}^{-10}, \left[\mathrm{salt}\right] = \left[\mathrm{base}\right] \\ \mathrm{pOH} =-\log K_{\mathrm{b}} + \log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{base}\right]} \\ \mathrm{pOH} = -\log \left(1\times {10}^{-10}\right) + \log 1 =10 \\ \mathrm{pH} + \mathrm{pOH} = 14 [\because \mathrm{concentration} \mathrm{of} [B^{-}] = \left[\mathrm{HB}\right] \\ \mathrm{pH} = 14-10 =4 \end{gathered}$$

(a) Higher is ox. no. of central atom in oxy-acid, e.g., acid (HClO4), more is its acidic nature and weaker is its conjugate base, (e.g., Cl${\mathrm{O}}_4^{-}$).

(b) Follow Le-Chatlelier's principle.

(b) A buffer of H₂CO₃ and ${\mathrm{HCO}}_3^{-}$ is formed.

(b) CH₃COOH is weak acid while NaOH is strong base, so one equivalent of CH₃COOH. Hence, one equivalent of each does not have pH value 7. As the NaOH is a strong base, the solution will be basic having a pH more than 7.

(a)    2AB₂$\rightleftharpoons$2AB(g) +B₂(g) 

           1          0         0

         1-x                x         x/2

         .^..             K_p = $\frac{{\mathit{x}}^{\mathit{2}}\mathit{.}\mathit{x}}{\mathit{2}\mathit{(}\mathit{1}\mathit{-}\mathit{x}\mathit{)}}{\left[\frac{\mathit{P}}{\mathit{1}\mathit{+}\mathit{x}\mathit{/}\mathit{2}}\right]}^{\mathit{1}}$

         x being small .^.. 1-x $\approx$1 and 1+ x/2 $\approx$ 1

          .^..  K_p = x³.P/2

     (d) Equilibrium constant, K=k_f/k_b 

                                 = Forward rate constant/ Backward rate constant

                                 = 1.1x10^-2/1.5x10^-3

                                 = 11/1.5 = 7.33

(a) Due to common ion effect, dissociation of NH₄OH decreases and thus, pOH increases or pH decreases.

(d) 2=[B]/[A]; 4=[C]/[B] and 6=[D]/[C]

     Thus, on multiplying

      2x4x6 = [D]/[A], i.e,  K_c for the reaction.

(d) These are characteristics of salt hydrolysis.