Question

Which of the following salts is the most basic in aqueous solution?

• A. ${\mathrm{CH}}_3\mathrm{COOK}$
• B. ${\mathrm{FeCl}}_3$
• C. $\mathrm{Pb}{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2$
• D. $\mathrm{Al}{\left(\mathrm{CN}\right)}_3$

Answer 1

The correct option is A

${\mathrm{CH}}_3\mathrm{COOK}$

Explanation for the correct option:

Option (a):

• The neutralization products of a weak acid and a strong base are called basic salt.
• Among the given options, option (a) is the most basic.
• This is because potassium acetate in an aqueous solution gives acetic acid (weak acid) and potassium hydroxide (strong base).
• $\underset{(\mathrm{Potassium}\mathrm{acetate})}{\underbrace{{\mathrm{CH}}_3\mathrm{COOK}\left(\mathrm{s}\right)}}+\underset{\left(\mathrm{Water}\right)}{\underbrace{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}}\to \underset{(\mathrm{Acetic}\mathrm{acid})}{\underbrace{{\mathrm{CH}}_3\mathrm{COOH}\left(\mathrm{aq}\right)}}+\underset{(\mathrm{Potassium}\mathrm{hydroxide})}{\underbrace{\mathrm{KOH}\left(\mathrm{aq}\right)}}$

Explanation for the incorrect option:

Option (b):

• Ferric chloride gives ferric hydroxide and hydrochloric acid in an aqueous medium.
• $\underset{(\mathrm{Ferric}\mathrm{chloride})}{\underbrace{{\mathrm{FeCl}}_3\left(\mathrm{s}\right)}}+\underset{\left(\mathrm{Water}\right)}{\underbrace{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}}\to \underset{(\mathrm{Ferric}\mathrm{oxide})}{\underbrace{\mathrm{Fe}{\left(\mathrm{OH}\right)}_3\left(\mathrm{aq}\right)}}+\underset{(\mathrm{Hydrochloric}\mathrm{acid})}{\underbrace{\mathrm{HCl}\left(\mathrm{aq}\right)}}$
• It is an acidic solution.

Option (c):

• The lead salt reacts with water to give a weak acid and a weak base.
• $\underset{(\mathrm{Lead}\mathrm{acetate})}{\underbrace{\mathrm{Pb}{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2\left(\mathrm{aq}\right)}}+\underset{\left(\mathrm{Water}\right)}{\underbrace{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}}\to \underset{\left(\mathrm{Acetate}\right)}{\underbrace{{\mathrm{CH}}_3\mathrm{COOH}\left(\mathrm{aq}\right)}}+\underset{(\mathrm{Lead}\mathrm{hydroxide})}{\underbrace{\mathrm{Pb}{\left(\mathrm{OH}\right)}_2\left(\mathrm{aq}\right)}}$

Option (d):

• The aluminum cyanide also forms a weak acid and a weak base with water.
• $\underset{(\mathrm{Alluminum}\mathrm{cyanide})}{\underbrace{\mathrm{Al}{\left(\mathrm{CN}\right)}_3\left(\mathrm{s}\right)}}+\underset{\left(\mathrm{Water}\right)}{\underbrace{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}}\to \underset{(\mathrm{Aluminum}\mathrm{hydroxide})}{\underbrace{\mathrm{Al}{\left(\mathrm{OH}\right)}_3\left(\mathrm{aq}\right)}}+\underset{\left(\mathrm{Hydrogen}\mathrm{cyanide}\right)}{\underbrace{\mathrm{HCN}\left(\mathrm{s}\right)}}$

Final answer:

Hence, Option (a) is the correct answer.