NEET Practice Questions (MCQs) with Answers & Solutions

(d)                      MX₂  $\to$         M²⁺   +   2X^-

Solubility 0.5 x 10^-4 M      0.5 x10^-4 M    2 x 0.5 x 10^-4 M

                                                      (on 100 % ionisation)

            .^.. K_sp of MX₂ = [M²⁺][X^-]²

                                = (0.5x10^-4)(1.0x10^-4)²

                                =0.5x10^-12 =5x10^-13

(a) Given K_a (NH₄OH) = 1.77 x 10^-5

     NH₄OH $\rightleftharpoons$ N${\mathrm{H}}_4^{+}$ + OH-

        K_a = $\frac{\left[{\mathrm{NH}}_4^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{NH}}_4\mathrm{OH}\right]}$ = 1.77 x 10^-5                         ..... (i)

     Hydrolysis of NH₄Cl takes place as

    NH₄Cl + H₂O $\to$NH₄OH + HCl

or N${\mathrm{H}}_4^{+}$ + H₂O $\to$ NH₄OH + H⁺

   Hydrolyis constant K_h = $\frac{\left[{\mathrm{NH}}_4\mathrm{OH}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{NH}}_4^{+}\right]}$                      .......(ii)

or K_h = $\frac{\left[{\mathrm{NH}}_4\mathrm{OH}\right]\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{NH}}_4^{+}\right]\left[{\mathrm{OH}}^{-}\right]}$                                      ..........(iii)

    From Eq. (i) and (iii)

    K_h = K_w/K_{a                                         [  $\because$[H+][OH-] = Kw]}

    _{Kh = 10^-14/1.77 x 10^-5 = 5.65 x 10^-10}

(d) $∆$n = 1 for this change; Unit of K_p =(atm)^$∆$n;

     Unit of K_c = (mol litre^-1)^$∆$n

(d) By the use of Henderson's equation

 $$\begin{gathered} \mathrm{pH} = \mathrm{p}{K}_{\mathrm{a}} + {\log }_{10} \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]} \\ \mathrm{When}, \left[\mathrm{salt}\right] = \left[\mathrm{acid}\right] \\ \therefore \mathrm{pH} = {\mathrm{pK}}_{\mathrm{a}} \\ \because {\mathrm{pK}}_{\mathrm{a}}= 3.58, \mathrm{thus} \mathrm{at} \mathrm{this} \mathrm{state} \\ \mathrm{pH} = 3.58 \end{gathered}$$

So, acetoacetic acid $({\mathrm{pK}}_{\mathrm{a}}= 3.58)$ is best to use.

(c) H₂X $\rightleftharpoons$H⁺ + HX^- ($K_{{\mathit{a}}_{\mathit{1}}}$)

      HX^- $\rightleftharpoons$H⁺ + X^2-  ($K_{{\mathit{a}}_{\mathit{2}}}$)

In Ist equation hydrogen ion is formed from neutral molecule whereas in IInd equation it comes from negatively charged species. Due to negative charge removal of proton is difficult.

 So,      $K_{{\mathit{a}}_{\mathit{1}}}$> $K_{{\mathit{a}}_{\mathit{2}}}$

(d)  CH₃COOH + C₂H₅OH $\rightleftharpoons$CH₃COOC₂H₅  + H₂O

            1               1                    0                 0

         (1-2/3)       (1-2/3)             2/3              2/3

               .^.. K_c = (2/3 x2/3)/(1/3 x 1/3) =4

(b) If K_w = 10^-12, then [H⁺] for neutral scale = 10^-6 or pH =6; thus, pH 6.9 refers for alkaline nature.

(b) K_p = K_c(RT)^-1/2

     ^..^. $\mathit{∆}n$= -1/2

(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${\mathrm{O}}_4^{-}$).