D.

pH > 7

Since, on dissolving NaCN in de-ionised water, following reaction takes place.

CN^- + H₂O $\rightleftharpoons$ HCN + OH^-

$\because$ CN^- ions come from the salt of strong base and weak acid.

$\therefore$ The solution becomes basic and pH > 7.

A.

formation of soluble complex anions like [PbCl₃]^-

PbCl₂ reacts with HCl as follows-

PbCl₂ (s) + 2Cl^- $\stackrel{\mathrm{Cold}}{\to }$ [PbCl₃]^- (aq)

PbCl₂ (s) + 2Cl^- $\stackrel{\mathrm{Excess} \mathrm{of} \mathrm{HCl}}{\to }$ [PbCl₄]^-2 (aq)

Thus, addition of excess amount of Cl^- ions changes the PbCl₂ as a soluble complex of [PbCl₄]^-2. Hence, it becomes soluble.

D.

Methyl orange is the weak organic base. It is used in the titration of WB vs SA

$\underset{Unionized form Yellow}{MeOH} \leftrightharpoons \underset{Ionized form \left(Red\right)}{M{e}^{+} }+ O{H}^{-}$

In a basic medium, equilibrium lies in the backward direction and therefore it shows yellow colour.

In acidic medium, equilibrium shifts in forwarding direction and therefore, colour changes from yellow to red.

A.

The N-atoms present in the ring will have the same pK_a values (6.0), while N atom outside the ring will have different pKa value (pK_a > 7.4)

Therefore, two N-atoms inside the ring will remain in the unprotonated form in human blood because of their pK_a(6.0) < pH of blood (7.4), while the N-atom outside the ring will remain in protonated form because of its pK_a > pH of blood (7.4).

A.

BCl₃ and AlCl₃

Both BCl₃ and AlCl₃ are Lewis acids as both ‘B’ and ‘Al’ has vacant p-orbitals. SiCl₄ is also a Lewis acid as silicon atom has vacant 3d-orbital.