NEET Practice Questions (MCQs) with Answers & Solutions

Given, CH₃COOH $\rightleftharpoons$ CH3COO^- + H⁺

K_a = 1.5 x 10^-5 ..... (i)

HCN $\rightleftharpoons$ H⁺ + CN^-; K_a = 4.5 x 10^-10 .... (ii)

CN^- + CH₃COOH $\rightleftharpoons$ HCN + CH₃COO^-

On substraction Eq. (ii)  from Eq. (i), we get

CH₃COOH+CN^-  $\rightleftharpoons$ HCN + CH₃COO^-;

K = K_a/K_a1 = 1.5 x 10^-5 / 4.5 x 10^-10 = 10⁵/3 = 3.33 x 10⁴

While adding two equations, dissociation constants are multiplied and when subtracting the equations, dissociation constants are divided.

Alternative 

CH₃COOH $\rightleftharpoons$  CH₃COO^-+H⁺;Ka = 1.5 x 10^-5

K_a = [CH₃COO^-][H⁺] / [CH₃COOH] = 1.5 x 10^-5 ..... (i)

HCN  $\rightleftharpoons$ H⁺ + CN^-; Ka = 4.5 x 10^-10

K_{a =} [H⁺][CN^-] / [HCN] = 4.5x10^-10 ..... (ii)

CN^- + CH₃COOH $\rightleftharpoons$ HCN + CH₃COOH^-

K_c = [HCN][CH₃COO^-] /[CN^-][CH₃COOH] .... (iii)

From Eqs. (i), (ii) and (iii)

K_c = 1.5 x 10^-5 / 4.5 x 10^-10 = 3.33 x 10⁴

Key Idea: The concentration of solids taken to be unity.

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

.^.. K = [Fe³⁺][OH^-]³

Hence, if OH^- ion concentration is decreased by 1/4 times, then equilibrium concentration of Fe³⁺ will increase by 64 times.

Key Idea: As the basic nature increases, pH increases

              pH of base>7

             pH of acid<7

In alkaline earth metals on moving downward the size of cation increases, thus basicity increases. Hence, the increasing order of basicity is as:

          MgCl₂<CaCl₂<SnCl₂<BaCl₂

Therefore, the solution of BaCl₂ will record the highest pH.

(b)

$$\begin{gathered} 2{\mathrm{AB}}_2\left(\mathrm{g}\right) \stackrel{ }{\rightleftharpoons }2\mathrm{AB}\left(\mathrm{g}\right) +\hspace{0.17em}{\mathrm{B}}_2\left(\mathrm{g}\right) \\ \mathrm{Initial} \mathrm{moles} 1 0 0 \\ \mathrm{At} \mathrm{equ}. 2(1-\mathrm{x}) 2\mathrm{x} \mathrm{x} \\ \mathrm{Where}, \mathrm{x}=\mathrm{degree} \mathrm{of} \mathrm{dissociation} \\ \mathrm{Total} \mathrm{moles} \mathrm{at} \mathrm{equilibrium} = 2-2\mathrm{x}+2\mathrm{x}+\mathrm{x} \\ = (2+\mathrm{x}) \\ \mathrm{So}, {\mathrm{P}}_{{\mathrm{AB}}_2} = \frac{2\left(1-\mathrm{x}\right)\mathrm{p}}{(2+\mathrm{x})} \\ {\mathrm{P}}_{\mathrm{AB}} = \frac{2\mathrm{xp}}{(2+\mathrm{x})} \\ {\mathrm{P}}_{{\mathrm{B}}_2} = \frac{\mathrm{xp}}{(2+\mathrm{x})} \\ {\mathrm{K}}_{\mathrm{p}} = \frac{{\left({\mathrm{P}}_{\mathrm{AB}}\right)}^2\left({\mathrm{P}}_{{\mathrm{B}}_2}\right)}{\left({\mathrm{P}}_{{\mathrm{AB}}_2}\right)} \\ = \frac{{\left({\displaystyle \frac{2\mathrm{xp}}{2+\mathrm{x}}}\right)}^2\left({\displaystyle \frac{\mathrm{x}}{2+\mathrm{x}}}\mathrm{p}\right)}{\left({\displaystyle \frac{2\left(1-\mathrm{x}\right)}{(2+\mathrm{x})}}\mathrm{p}\right)} \\ = \frac{{\mathrm{x}}^3\mathrm{p}}{\left(2+\mathrm{x}\right)(1-\mathrm{x}{)}^2} \\ \left[\because \mathrm{x}<<<1 \mathrm{and} 2, \mathrm{so} (1-\mathrm{x})\approx 1, (2+\mathrm{x})\approx 2\right] \\ =\frac{{\mathrm{x}}^3\mathrm{p}}{2} \\ \mathrm{x}={\left(\frac{2{\mathrm{K}}_{\mathrm{p}}}{\mathrm{p}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \end{gathered}$$

From Equation,

       X$\rightleftharpoons$Y + Z

       1     0     0 Initial mole

  (1-$\mathrm{\alpha }$)    $\mathrm{\alpha }$     $\mathrm{\alpha }$

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

             K₁ = [H₂]^1/2[I₂]^1/2/[HI]                   ....(i)

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

              K₂ = [HI]²/[H₂][I₂]                          ....(ii)

From eqs (i) and (ii)

                ${\mathrm{K}}_1^2$= 1/K₂

._..            K₁ = 8.0

.^..            K₂ = 1/(K₁)² = 1/8² = 1/64

[H₃O⁺] = [H⁺] = 10^-10

               pH +pOH = 14

                pH = -log[H⁺]

                pH = -log[10^-10]

               pH=10

             pOH + 10 =14

             pOH = 14-10 =4

HA $\rightleftharpoons$H⁺ + A^-

K_a = [H⁺][A⁺]/[HA] = [H⁺]²/[HA]

[H⁺] = $\sqrt{{\mathrm{K}}_{\mathrm{a}}\left[\mathrm{HA}\right] }=\sqrt{1\mathrm{x}{10}^{-5}\mathrm{x}0.1}=\sqrt{1\mathrm{x}{10}^{-6}}$ = 1x10^-3

$\mathrm{\alpha }$ = actual ionization / molar concentration = 10^-3 / 0.1 = 10^-2

% of acid dissociated = 10^-2 x 100 = 1%

Fluorine is the most electronegative element in the periodic table. So, it has the greatest proton affinity to form stable compounds.

Cu (s) + 2Ag²⁺ (aq) $\to$Cu²⁺ (aq) + 2Ag (s)

                E$°$ = 0.46 V at 298 K

             RT ln K = nFE$°$

                 ln K = nFE$°$/RT = (2 x 0.46)/0.0591 

                    K = 4 x 10¹⁵