Considering the tube wherein NaOH (1 M) was added to benzoic acid.
a) what reaction would occur in the tube?
b) write a balance equation that supports your explanation above?
c) describe what would happen if u added HCL (6 M) to the same tube?
please help!! thank you
Considering the tube wherein NaOH (1 M) was added to benzoic acid.
a) what reaction would occur in the tube?
b) write a balance equation that supports your explanation above?
c) describe what would happen if u added HCL (6 M) to the same tube?
please help!! thank you
Answered above.
a) In the tube, the reaction that would occur is a neutralization reaction between sodium hydroxide (NaOH) and benzoic acid. Benzoic acid is a weak acid, while sodium hydroxide is a strong base. When the two compounds come into contact, they will react to form water (H2O) and a salt called sodium benzoate.
b) The balanced equation for the neutralization reaction between NaOH and benzoic acid can be represented as follows:
NaOH + C6H5COOH -> C6H5COONa + H2O
This equation shows that one molecule of sodium hydroxide reacts with one molecule of benzoic acid to produce one molecule of sodium benzoate and one molecule of water.
c) If hydrochloric acid (HCl) is added to the same tube, a new reaction would occur. This time, the HCl would react with the sodium benzoate that was formed in the previous step. This reaction is called an acid-base reaction or a proton transfer reaction. The hydrochloric acid is a strong acid, and it will donate a hydrogen ion (H+) to the sodium benzoate, forming benzoic acid and sodium chloride.
The balanced equation for the reaction between HCl and sodium benzoate can be represented as follows:
C6H5COONa + HCl -> C6H5COOH + NaCl
This equation shows that one molecule of sodium benzoate reacts with one molecule of hydrochloric acid to produce one molecule of benzoic acid and one molecule of sodium chloride.