Question

A colourless mixture of two substances (A) and (B) (excess) is soluble in $H_{2}O$, separately. (A) turns blue litmus red and (B) turns red litmus blue. (A) gives white ppt. with (B), which dissolves in excess of (B) forming (C). (A) when placed in moist air gives fumes and can form dimer. (A) gives white ppt. with ${NH}_{4}Cl$ and ${NH}_{4}OH$ soluble in (B). (A) also gives white precipitate with ${AgNO}_3$ soluble in ${NH}_{4}OH$.

Thus, A turns blue litmus red. It is due to acidic solution because of ________.

• A. Hydrolysis of cation of A
• B. Hydrolysis of anion of A
• C. Hydrolysis of both cation and anion
• D. None of the above is correct

Answer 1

The correct option is A Hydrolysis of cation of A

The colourless mixture contains ${AlCl}_3$ as A and $NaOH$ as B which is present in excess.

${AlCl}_3$ is acidic in nature. It turns blue litmus red.

$NaOH$ is alkaline in nature. It turns red litmus blue.

${AlCl}_3$ reacts with $NaOH$ to give a white ppt of $Al(OH{)}_3$ which is soluble in more $NaOH$ to form $NaAl(OH{)}_4$

${AlCl}_3+3NaOH\to Al(OH{)}_3\downarrow (white)+3NaCl$

$Al(OH{)}_3+NaOH\to NaAl(OH{)}_4$

With moist air, ${AlCl}_3$ gives fumes of HCl.

${AlCl}_3+3H_{2}O\to Al(OH{)}_3+3HCl$

Therefore, option A is correct.