Chemistry MCQs for NEET — Practice Questions with Answers

For Mx, K_sp =S²

     For MX_2, K_sp =4S³

     For M₃X, K_sp =27S ⁴

Use the above formula and calculate the solubilities.

 $$\begin{gathered} \mathrm{Solubility} \left(\mathrm{S}\right)=\frac{\mathrm{K}\times 1000}{{\mathrm{\Lambda }}_{\mathrm{m}}^{\infty }} \\ =\frac{2.3\times {10}^{-6}\times 1000}{61.9 + 76.3} \\ =\frac{2.3\times {10}^{-3}}{138.2} \\ =1.66 \times {10}^{-5}\mathrm{mol}/\mathrm{L} \end{gathered}$$

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

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