Which of the following substances, when added to a solution of nitrous acid [tex]$\left( HNO_2\right)$[/tex], could be used to prepare a buffer solution?

A) [tex]$HC_2H_3O_2$[/tex]
B) [tex]$NaCl$[/tex]
C) [tex]$HCl$[/tex]
D) [tex]$NaNO_2$[/tex]



Answer :

To prepare a buffer solution, you need a combination of a weak acid and its conjugate base. A buffer solution resists changes in pH when small amounts of acid or base are added.

In this scenario, we have nitrous acid [tex]$\left( HNO _2 \right)$[/tex], which is a weak acid. To form a buffer, we need its conjugate base, which in the case of nitrous acid is the nitrite ion [tex]$\left( NO _2^{-} \right)$[/tex]. Let's analyze the given options one by one to determine which one can provide the needed conjugate base.

A) [tex]$\mathbf{HC _2 H _3 O _2}$[/tex] (Acetic acid): This is another weak acid, but it does not provide the conjugate base of nitrous acid. Instead, it breaks down into acetate ions ([tex]$CH _3 CO _2^{-}$[/tex]) in solution, which are not useful for our purpose here.

B) [tex]$\mathbf{NaCl}$[/tex] (Sodium chloride): This compound dissociates into sodium ions ([tex]$Na^{+}$[/tex]) and chloride ions ([tex]$Cl^{-}$[/tex]) in solution. Neither of these ions forms the conjugate base of nitrous acid, so sodium chloride cannot be used to make a buffer with [tex]$HNO _2$[/tex].

C) [tex]$\mathbf{HCl}$[/tex] (Hydrochloric acid): Hydrochloric acid is a strong acid that dissociates completely in solution. Addition of a strong acid would not provide the conjugate base of nitrous acid. In fact, it would likely lower the pH significantly and disrupt any buffer system.

D) [tex]$\mathbf{NaNO _2}$[/tex] (Sodium nitrite): Sodium nitrite dissociates into sodium ions ([tex]$Na^{+}$[/tex]) and nitrite ions [tex]$\left( NO _2^{-} \right)$[/tex]. The nitrite ion is the conjugate base of nitrous acid [tex]$\left( HNO _2 \right)$[/tex]. Therefore, adding sodium nitrite to a nitrous acid solution introduces the necessary conjugate base to create a buffer solution.

Based on this analysis, the correct answer is:

D) [tex]$\mathbf{NaNO _2}$[/tex]