Chemistry MCQs for NEET — Practice Questions with Answers

(b) Given, pH of Ba(OH)₂ = 12.

.^.. pOH = 14-pH = 14-12 = 2

We know that,  pOH = -log[OH]^-

                           2= -log[OH]^-

                       [OH]^- = antilog(-2)

                       [OH]^- = 1 x 10^-2 

Ba(OH)² dissolves in water as:

 Ba(OH)² (s) $\rightleftharpoons$ Ba²⁺ + 2OH^-

 S mol L-1           S        2S

.^.. [OH]^- = 2S = 1 x 10^-2

S=[OH]^-/2                             [Ba²⁺ =S]

Ba²⁺ = [OH]^-/2 = (1 x 10^-2)/2

K_sp = [Ba²⁺][OH^-]²

      =[(1 x10^-2)/2](1 x 10^-2)2

      = 0.5 x 10^-6 = 5 x 10^-7

(c) $K_{{\mathit{c}}_{\mathit{1}}}$ = 1/$K_{c_{\mathit{2}}}$ = 1/(2.4 x 10^-3) = 4.2 x 10²

(c) 2HI $\rightleftharpoons$H₂ + I₂;

           Kc = $\frac{{\alpha }^2}{4(1-\alpha {)}^2}$

where $\alpha$ is degree of dissociation,

   Also,         $\alpha$ =22/100

  .^.. K_c = 0.0199

(c)          N₂ + 3H₂ $\rightleftharpoons$ 2NH_{3                                                                                                                                        28/28 = 1    6/2=3          0             mole before reaction                                                                                                    1-1/2         3-3/2        17/17 =1  mole after reaction}

        .^.. Mole of N₂ = 1/2

        .^.. mass of N₂ = 14 g

           Mole of H₂ = 3/2

        .^.. mass of H₂ = 3/2 x 2 = 3 g

The strength of an acid depends on its ability to dissociate and release H+ ions. Among the given options, HClO4 (perchloric acid) is the strongest acid due to the high electronegativity of the central chlorine atom, which stabilizes the negative charge on the anion effectively.

(a) CH₃COO- + H₂O $\rightleftharpoons$CH₃COOH + OH^-

     ...                [OH^-] = c.h = $c\sqrt{\frac{K_H}{c}}=\sqrt{K_H.c}= \sqrt{\frac{K_w.c}{K_a}}$

      or               -log OH =  $\frac{1}{2}$[logK_w + log c- logK_a]

      or                 pOH =  $\frac{1}{2}$[pK_w -log c - pK_a]

 Now, pH + pOH = pKw

                  pH=  $\frac{1}{2}$[pK_w + log c + pK_a]

(c)    Fe(OH)₃(s) $\rightleftharpoons$Fe³⁺(aq) + 3OH^- (aq)

                         K =[Fe³⁺][OH^-]³/Fe(OH)³]            ....(i)

To maintain equilibrium constant, let the concentration of Fe³⁺ be increased by x times on decreasing the concentration of OH^- by 1/4 times.

                        K=[xFe³⁺][1/4 x OH^-]³/ [Fe(OH)₃]      .....(ii)

  From Eqs. (i) and (ii)

                           1/64 x x=1

                                    x = 64 times

(b) HI(g) $\rightleftharpoons$1/2 H₂(g) + 1/2 I₂(g)

                        K= [I₂]^1/2[H₂]^1/2/[HI]     .              ........(i)

    H₂(g) + I₂(g) $\rightleftharpoons$2HI

                        K' = [HI]²/[H₂][I₂]                          .....(ii)

          From Eqs. (i) and (ii)

                  K x $\sqrt{K\text{'}}$=1

                       K' =1/K² = 1/8² =1/64

The combination of a weak acid (CH3COOH) and its salt (CH3COONa) can act as an acidic buffer solution. When CH3COOH dissociates, it produces H+ ions, and CH3COONa provides a reserve of CH3COO- ions to neutralize any added base, maintaining a relatively constant pH.

(d) Dissociation of weak acid decreases in presence of common ion.

CH₃COOH(l) + H₂O(l) <——-> H₃O⁺(aq) + CH₃COO^-(aq)

HCl + H2O ——-> H₃O⁺(aq) + Cl^-(aq)

(d) All are binary salts, hence their solubility is equal to square root of their solubility products. So, order of solubility is CuS>CdS>HgS or greater the value of solubility product, greater will be the solubility.