Formic acid is only readily soluble in water and in both 0.6 M NaHCO3 and 2.5 M NaOH.
give a brief explanation of the reaction for each
Same answer as above for hexanoic acid.
These are typical reactions between and acid and a bicarbonate, and an acid and a base. Sodium formate (a salt) and H2O are forming in both cases. CO2 is forming typically when a base reacts with a bicarbonate or a carbonate.
HCOOH + NaHCO3 ---> HCOONa + H2O + CO2
HCOOH + NaOH ---> HCOONa + H2O
Sure! Let's break down the reactions for the solubility of formic acid in water, 0.6 M NaHCO3 (sodium bicarbonate), and 2.5 M NaOH (sodium hydroxide).
1. Solubility in water:
When formic acid (HCOOH) dissolves in water (H2O), it undergoes a reversible reaction known as hydrogen bonding. The formic acid molecules dissociate into hydrogen ions (H+) and formate ions (HCOO-). The reaction can be represented as follows:
HCOOH + H2O β H3O+ + HCOO-
2. Solubility in 0.6 M NaHCO3:
In this case, the sodium bicarbonate (NaHCO3) dissociates in water to release sodium ions (Na+) and bicarbonate ions (HCO3-). When formic acid is added, it reacts with the bicarbonate ions in a neutralization reaction, forming sodium formate (NaHCOO) and releasing carbon dioxide (CO2) gas. The reaction can be shown as:
HCOOH + HCO3- β HCOO- + H2O + CO2
3. Solubility in 2.5 M NaOH:
Similarly, sodium hydroxide (NaOH) dissociates in water to form sodium ions (Na+) and hydroxide ions (OH-). When formic acid is added, it reacts with the hydroxide ions in a neutralization reaction, forming sodium formate (NaHCOO) and water (H2O). The reaction can be represented as:
HCOOH + OH- β HCOO- + H2O
It's important to note that in both the NaHCO3 and NaOH reactions, formic acid is neutralized and converted into its conjugate base, formate ion (HCOO-), which increases its solubility.