Question

The pH of 0.1 M solution of the following salts increases in order

• A. $NaCl$ < ${NH}_{4}Cl$ < $NaCN$ < $HCl$
• B. $NaCN$ < ${NH}_{4}Cl$ < $NaCl$ < $HCl$
• C. $HCl$ < $NaCl$ < $NaCN$ < ${NH}_{4}Cl$
• D. $HCl$ < ${NH}_{4}Cl$ < $NaCl$ < $NaCN$

Answer 1

The correct option is D

$\mathrm{HCl}$ < ${\mathrm{NH}}_4\mathrm{Cl}$ < $\mathrm{NaCl}$ < $\mathrm{NaCN}$

The explanation for the correct option (D):

• Salt is a chemical compound that is produced by replacing the Hydrogen ($\mathrm{H}$) in acid with a metal.
• For example, Hydrochloric acid ($\mathrm{HCl}$) is an acid. If $\mathrm{H}$ of $\mathrm{HCl}$ is replaced by a metal, Sodium ($\mathrm{Na}$), then a Sodium chloride ($\mathrm{NaCl}$) salt is formed.
• The pH of salt solutions can be determined by the hydrolysis of salts.
• The salts of strong bases and strong acids produce neutral solutions.
• The salts of strong acids and weak bases produce acidic solutions.
• The salts of strong bases and weak acids produce basic solutions.
• When the pH of the solution is less than 7, the solution is said to be acidic. When the pH of the solution is above 7, the solution is basic and when it is equal to 7, the solution is neutral.
• Sodium chloride ($\mathrm{NaCl}$) is a salt of a strong base, Sodium hydroxide ($\mathrm{NaOH}$), and a strong acid, $\mathrm{HCl}$.
• $$\begin{gathered} \mathrm{NaCl}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to \mathrm{NaOH}\left(\mathrm{aq}\right)+\mathrm{HCl}\left(\mathrm{aq}\right) \\ \mathrm{Water} \end{gathered}$$
• Hence, $\mathrm{NaCl}$ is neutral in nature and has a pH of 7.
• Sodium cyanide ($\mathrm{NaCN}$) is a salt of a strong base, Sodium hydroxide ($\mathrm{NaOH}$), and a weak acid, Hydrogen cyanide ($\mathrm{HCN}$).
• $\mathrm{NaCN}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to \mathrm{NaOH}\left(\mathrm{aq}\right)+\mathrm{HCN}\left(\mathrm{aq}\right)$
• Hence, $\mathrm{NaCN}$ is basic in nature and has a pH greater than 7.
• Ammonium chloride (${\mathrm{NH}}_4\mathrm{Cl}$) is a salt of a weak base, Ammonium hydroxide (${\mathrm{NH}}_4\mathrm{OH}$), and a strong acid, $\mathrm{HCl}$.
• ${\mathrm{NH}}_4\mathrm{Cl}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{NH}}_4\mathrm{OH}\left(\mathrm{aq}\right)+\mathrm{HCl}\left(\mathrm{aq}\right)$
• Hence, ${\mathrm{NH}}_4\mathrm{Cl}$ is acidic in nature and has a pH of less than 7.
• $\mathrm{HCl}$ is a strong acid as it completely ionizes in water to produce Hydrogen (${\mathrm{H}}^{+}$) ions.
• $$\begin{gathered} \mathrm{HCl}\left(\mathrm{aq}\right)\to {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right) \\ \mathrm{Chloride}\mathrm{ion} \end{gathered}$$
• Hence $\mathrm{HCl}$ is strongly acidic and has the lowest pH among the given solutions.
• Therefore, $\mathrm{NaCN}$ has the highest pH followed by $\mathrm{NaCl}$, then ${\mathrm{NH}}_4\mathrm{Cl}$ and $\mathrm{HCl}$ has the lowest pH.
• $\mathrm{HCl}$ < ${\mathrm{NH}}_4\mathrm{Cl}$ < $\mathrm{NaCl}$ < $\mathrm{NaCN}$

Final Answer: Hence, option (D) is correct. The pH of 0.1 M solution of the following salts increases in order $\mathrm{HCl}$ < ${\mathrm{NH}}_4\mathrm{Cl}$ < $\mathrm{NaCl}$ < $\mathrm{NaCN}$.