5 millimoles of caustic potash and 5 millimoles of oxalic acid are mixed and dissolved in 100 mL water. The solution will be

• basic

• acidic

• neutral

• cannot say

Which solution is a buffer?

• Acetic acid + NaOH (equimolar ratio)

• Acetic acid + NaOH (1 : 2 molar ratio)

• Acetic acid + NaOH (2 : 1 molar ratio)

• HCl + NaOH (equimolar ratio)

The solubility of Al(OH)₃ is 'S' mol L^-1. The solubility product will be

• S²

• S³

• 27 S⁴

• 27 S³

If in the reaction N₂O₄ $\rightleftharpoons$2NO₂: $\mathrm{\alpha }$ is the degree of dissociation of N₂O₄ then total number of moles at equilibrium is

• (1-α)

• (1+α)

• (1-α)²

• (1+α)²

$\frac{{\mathrm{K}}_{\mathrm{p}}}{{\mathrm{K}}_{\mathrm{c}}}$ for the reaction

CO (g) + $\frac{1}{2}$ O₂ (g) $\rightleftharpoons$ CO₂ (g) is

• RT

• $\frac{1}{\sqrt{\mathrm{RT}}}$

• $\sqrt{\mathrm{RT}}$

• $\frac{1}{\mathrm{RT}}$

In the equilibrium mixture, KI+ I₂ $\rightleftharpoons$KI₃ the concentration of KI and I₂ and three fold respectively . The concentration of KI₃ becomes

• two fold

• three fold

• five fold

• six fold

Consider the following reversible reactions,

N₂(g) + 3H₂(g) $\rightleftharpoons$ 2NH₃(g) (K₁)  ...(i)

N₂ (g) + O₂(g) $\rightleftharpoons$2NO (g) (K₂) ....(ii)

H₂ (g) + $\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right) \rightleftharpoons {\mathrm{H}}_2\mathrm{O} \left(\mathrm{g}\right) \left({\mathrm{K}}_3\right) ....\left(\mathrm{iii}\right)$

The equilibrium constant for the reaction:

2NH₃(g) + $\frac{5}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\hspace{0.17em}\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right) + 3{\mathrm{H}}_2\mathrm{O} \left(\mathrm{g}\right)$ will be

• K₁ K₂ K₃

• $\frac{{\mathrm{K}}_1{\mathrm{K}}_2}{{\mathrm{K}}_3}$

• $\frac{{\mathrm{K}}_1{\mathrm{K}}_3^3}{{\mathrm{K}}_2}$

• $\frac{{\mathrm{K}}_2{\mathrm{K}}_3^3}{{\mathrm{K}}_1}$

618.

The number of moles of sodium acetate to be added to 0.1 M acetic acid for the buffer to have a pH = 4.7 is [pK_a for acetic acid is 4.7]

 

• 0.2

• 0.4

• 0.1M

• None of these

If K_c is the equilibrium constant for the formation of NH₃, the dissociation constant of NH₃ under the same condition will be

• $\frac{1}{{\mathrm{K}}_{\mathrm{c}}}$

• ${\mathrm{K}}_{\mathrm{C}}^2$

• $\sqrt{{\mathrm{K}}_{\mathrm{C}}}$

• K_C

The pH of a soft drink is 3.92. The hydrogen ion concentration will be

( given antilog 0.08 = 1.2)

• 1.96 × 10^-2 mol L^-1

• 1.96 × 10^-3 mol L^-1

• 1.2 × 10^-4 mol L^-1

• 1.2 × 10^-3 mol L^-1