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?
Have you changed compounds. Your earlier question concerned 4-aminoethyl benzoate. Straight benzoic acid would form a salt since it is an acid reacting with a base, NaOH.
ArCOOH + NaOH ==> H2O + ArCOONa where Ar is the benzene ring.
c). If HCl is added to the tube, HCl neutralizes the NaOH and reprecipirates the benzoic acid.
a) In the tube where NaOH (1 M) was added to benzoic acid, a neutralization reaction would occur. Benzoic acid is an organic acid and NaOH is a strong base. When they react, the benzoic acid (acid) would donate a hydrogen ion (H+) to NaOH (base), forming sodium benzoate (salt).
b) The balanced equation for the neutralization reaction between benzoic acid and NaOH is as follows:
C6H5COOH + NaOH -> C6H5COONa + H2O
c) If HCl (6 M) is added to the same tube containing sodium benzoate, a new reaction would occur. HCl is a strong acid, and sodium benzoate is the salt formed from the previous reaction. When they react, the hydrogen ions (H+) from HCl would displace the sodium ions (Na+) from sodium benzoate, resulting in the formation of benzoic acid and sodium chloride.
The balanced equation for this reaction is:
C6H5COONa + HCl -> C6H5COOH + NaCl