Equilibrium MCQs for NEET — Chemistry Questions with Answers

(b) Key Idea As solubility of AgCl(s) is asked in 0.1 M NaCl, so in the calculation, solubility of C^l- (from NaCl) must be added to the solubility of Cl^- (from AgCl).

Let s be the solubility of Ag⁺ and Cl^- in AgCl before the addition of NaCl.

         NaCl (aq) $\to$Na⁺ (aq) + Cl^-(aq)

          0.1 M          0                0

           0               0.1 M         0.1+s

       AgCl(s) $\rightleftharpoons$Ag⁺(aq) + Cl^-(aq)

                         s             s+0.1

Given,           K_sp = 1.6 x 10^-10 = [Ag⁺][Cl^-]

        or                   1.6 x 10^-10 = s(0.1+s)

                                                = 0.1s + s²

._.. K_sp is small, so s very less in comparison with 0.1. Hence s² can be neglected.

         Thus, 1.6 x 10-10 = 0.1s

   or                          s = 1.6 x 10-9 M

(c) Boric acid can be considered as an acid because its molecule accepts OH' from water, releasing proton.

H₃BO₃ + H₂O $\rightleftharpoons$   B(OH)^-₄   +   H⁺

  Acid     Base     Conjugate  Conjugate

                            base           acid

Remember In the given options to the question, (a), (b) and (c) are correct as all of these sentences have more or less similar meaning but here (c) option is the most appropriate one as it gives complete explanation of the fact that how boric acid can be combined with an acid.

Among the given options, PF3 (phosphorus trifluoride) is the most likely to behave as a Lewis base. It has an available lone pair of electrons on the phosphorus atom, which can be donated to an electron-deficient Lewis acid, forming a coordinate covalent bond.

(a) As the oxidation state of halogen i.e. —Cl in this case increases, acidity of oxyacid increases.

HClO : Oxidation state of Cl = +1

HClO₂: Oxidation state of Cl = + 3

HClO₂; Oxidation state of Cl = + 5

HClO₄ Oxidation state of Cl = + 7

Therefore, the correct order of acidity would be

HClO < HClO₂ < HClO₃ < HClO₄

For a reaction,

$$\begin{gathered} \underset{\mathrm{Reactant}}{\mathrm{A}}\rightleftharpoons \underset{\mathrm{Product}}{\mathrm{B}} \\ \mathrm{K}=\frac{{\left[\mathrm{B}\right]}_{\mathrm{eq}}}{{\left[\mathrm{A}\right]}_{\mathrm{eq}}}\Rightarrow 1.6\times {10}^{12}=\frac{{\left[\mathrm{B}\right]}_{\mathrm{eq}}}{{\left[\mathrm{A}\right]}_{\mathrm{eq}}} \\ {\left[\mathrm{B}\right]}_{\mathrm{eq}}>>{\left[\mathrm{A}\right]}_{\mathrm{eq}} \end{gathered}$$

So, mostly the product will be present in te equilibrium mixture.

Mathematical expression of the thermodynamic equilibrium is

ΔG = ΔG° + 2.303 RT logQ

At equilibrium when ΔG = 0 and O = K

then ΔG = ΔG° + 2.303 RT logK = 0

ΔG° = - 2.303 RT log K

(a) Key Concept When equal volumes of acid and base are mixed, then resulting solution become alkaline if concentration of base is taken high.

Let normality of the solution after mixing 0.1 M NaOH and 0.01 M HCl is N.

$$\begin{gathered} \therefore \mathrm{N}1\mathrm{V}1-\mathrm{N}2\mathrm{V}2 = \mathrm{NV} \\ \mathrm{or} 0.1 \times 1 - 0.01 \times 1=\mathrm{N}\times 2 \end{gathered}$$

Since, normality of NaOH is more than that of HCl. Hence, the resulting solution is alkaline.

$$\begin{gathered} \mathrm{or} \left[\mathrm{OH}\right]=\mathrm{N}=\frac{0.09}{2}=0.045\mathrm{N} \\ \mathrm{or} \mathrm{pOH} =-\log (0.045)=1.35 \\ \therefore \mathrm{pH} =14-\mathrm{pOH} \\ =14-1.35=12.65 \end{gathered}$$