Write the reactions that occur for the benzoic acid/sodium benzoate buffer when you react it with HNO3 and KOH. How would I go about writing it? Thanks.
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To write the reactions that occur for the benzoic acid/sodium benzoate buffer when reacting with HNO3 and KOH, you need to understand the properties and behavior of these compounds.
First, let's consider the benzoic acid/sodium benzoate buffer. This buffer system consists of a weak acid, benzoic acid (C6H5COOH), and its conjugate base, sodium benzoate (C6H5COONa). Buffers are used to resist changes in pH when small amounts of acid or base are added.
Now, let's consider the reactions with HNO3 (nitric acid) and KOH (potassium hydroxide).
1. Reaction with HNO3:
HNO3 is a strong acid that can donate a proton (H+) to the benzoic acid. The acidic medium favors the transfer of protons. In this case, the following reaction occurs:
C6H5COOH (benzoic acid) + HNO3 → C6H5COOH (protonated benzoic acid) + NO3- (nitrate ion)
Note: The protonated benzoic acid is formed when HNO3 donates a proton to the benzoic acid, resulting in the formation of a positively charged species.
2. Reaction with KOH:
KOH is a strong base that can accept a proton. The basic medium favors the transfer of protons. In this case, the following reaction occurs:
C6H5COONa (sodium benzoate) + KOH → C6H5COOH (benzoic acid) + NaOH (sodium hydroxide)
Note: The sodium ion (Na+) from sodium benzoate remains unchanged, while the hydroxide ion (OH-) from KOH reacts with the benzoate ion to form benzoic acid and sodium hydroxide.
To summarize, when reacting the benzoic acid/sodium benzoate buffer with HNO3, the benzoic acid gets protonated, and nitrate ions are formed. When reacting with KOH, the sodium benzoate gets transformed into benzoic acid and sodium hydroxide.
Remember to always consider the basic concepts of acid-base reactions, the nature of the compounds involved, and the medium in which the reactions occur.
Let's call benzoic acid HBz and the benzoate ion then is Bz^-
With HNO3, the added acid (HNO3) reacts with the base, Bz^-.
Bz^- + H^+ ==> HBz
With KOH, the added base (KOH) reacts with the acid, HBz.
HBz + OH^- ==> Bz^- + H2O