Equilibrium MCQs for NEET — Chemistry Questions with Answers

In presence of common ion, solubility of salt decreases.

For pH=1,     N=10-1

For pH =2,    N=10-2

N₁V₁ + N₂V₂ = NV

10^-1 X 50 + 10^-2 X 100 = N X 150

N=4 X 10^-2

pH =-log[H⁺] = -log4 X 10^-2

pH=1.4

The sulfuric acid in this mixture is sufficiently strong to protonate nitric acid, producing the nitronium ion (NO₂⁺), which is the active species.

2H₂SO₄ + HNO₃ → NO₂⁺ + H₃O⁺ + 2HSO₄^-

NaCl → Na+ + Cl−

H2O → H+ + OH−

Hydrogen & chloride ions discharge at negative & positive electrodes respectively,leaving behind sodium & hydroxide ions.These then combine to form sodium hydroxide,a strong base,

For half neutralisation of weak acid from strong base,

pH=pK_a

4.2=-log K_a

K_a=antilog of -4.2=6.31 X 10^-5 

due to least tendency to form BACK BONDING by Br atom.

pH=8, pOH = 6, [OH^-] = 10^-6

$$\begin{gathered} \mathrm{Fe}(\mathrm{OH}{)}_2\rightleftharpoons {\mathrm{Fe}}_{}^{2+} + 2{\mathrm{OH}}^{-} \\ \mathrm{x} 2\mathrm{x} \\ {\mathrm{K}}_{\mathrm{sp}} = \mathrm{x} \times [{\mathrm{OH}}^{-}{]}^2 \\ \mathrm{x}=\frac{1.6 \times {10}^{-14}}{({10}^{-6}{)}^2}=0.016 \end{gathered}$$

Use the formula of salt hydrolysis of salt of WA and SB.

$\mathrm{pH}=-\frac{1}{2}{\mathrm{logK}}_{\mathrm{w}}-\frac{1}{2}\log {\mathrm{K}}_{\mathrm{a}}+\frac{1}{2}\mathrm{logC}$

(a) For the reaction,

SrCO₃(s)$\rightleftharpoons$ SrO(s) + CO₂(q),

Kp = 1.6 atm = p_CO2 = maximum pressure of CO₂

Given, p, = 0.4 atm, V₁=  20 L, T₁ = 400 K

P₂ = 1.6 atm. V₂ =?. T₂ =400 K

At constant temperature, p₁V₁ = p₂V₂

0.4x20=1.6 xV₂

V₂=(0.4x20)/1.6=5L

(b)

$$\begin{gathered} \mathrm{Given}, {\mathrm{N}}_2 + 3{\mathrm{H}}_2 \stackrel{ }{\rightleftharpoons }2{\mathrm{NH}}_3, {\mathrm{K}}_1 ...\left(\mathrm{i}\right) \\ {\mathrm{N}}_2 + {\mathrm{O}}_2 \stackrel{ }{\rightleftharpoons }2\mathrm{NO}, {\mathrm{K}}_2 ...\left(\mathrm{ii}\right) \\ {\mathrm{N}}_2 + \frac{1}{2}{\mathrm{O}}_2 \stackrel{ }{\stackrel{ }{\to }}{\mathrm{H}}_2\mathrm{O}, {\mathrm{K}}_3 ...\left(\mathrm{iii}\right) \\ \mathrm{To} \mathrm{calculate}, \\ 2{\mathrm{NH}}_3 + \frac{5}{2}{\mathrm{O}}_2 \stackrel{ \mathrm{K} }{\rightleftharpoons }2\mathrm{NO} +\hspace{0.17em}3{\mathrm{H}}_2\mathrm{O}, \mathrm{K} = ? ...\left(\mathrm{iv}\right) \end{gathered}$$

On reversing the equation (i) and multiplying the equation (iii) by 3, we get

$$\begin{gathered} 2{\mathrm{NH}}_3\stackrel{ }{\rightleftharpoons }{\mathrm{N}}_2 +\hspace{0.17em}3{\mathrm{H}}_2, \frac{1}{{\mathrm{K}}_1} ...\left(\mathrm{v}\right) \\ 3{\mathrm{H}}_2 + \frac{3}{2}{\mathrm{O}}_2 \stackrel{ }{\to } 3{\mathrm{H}}_2\mathrm{O}, {\mathrm{K}}_3^3 ...\left(\mathrm{vi}\right) \\ \mathrm{Now}, \mathrm{add} \mathrm{equation}. \left(\mathrm{ii}\right), \left(\mathrm{v}\right) \mathrm{and} \left(\mathrm{vi}\right), \mathrm{we} \mathrm{get} \mathrm{the} \mathrm{resultant} \mathrm{equation}. \left(\mathrm{iv}\right). \\ 2{\mathrm{NH}}_3 + \frac{5}{2}{\mathrm{O}}_2 \stackrel{ \mathrm{K} }{\rightleftharpoons }2\mathrm{NO} + 3{\mathrm{H}}_2\mathrm{O} \\ \therefore \mathrm{K}=\frac{{\mathrm{K}}_2{\mathrm{K}}_3^3}{{\mathrm{K}}_1} \end{gathered}$$