NEET Practice Questions (MCQs) with Answers & Solutions

4. N₂O₄ $\rightleftharpoons$2NO₂

       1         0     ---- at zero times

     1-$\mathrm{\alpha }$       2$\mathrm{\alpha }$   ---- at equilibrium

 No. of moles at equilibrium = 1-$\mathrm{\alpha }$+2$\mathrm{\alpha }$= 1+$\mathrm{\alpha }$

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