We are asked to write the three products of the reaction between CH3COOH and NaHCO3(given one is a gas)
They would be,
1)CH3COO^-(Na)^+
2)CO2(g)
And what would be the third product?
As NaHCO3 is acidic would that be Na2CO3?
Because CH3COO^-(Na)^+ is a weak acid salt and NaHCO3 is acidic can we take
CH3COO^-(Na^+) + NaHCO3-->Na2CO3 + CH3COOH
Or can we take Sodium acetate will take H^+ from water and become CH3COOH,as it is a weak acid(sodium acetate)?
Thank you!
#1. NaHCO3 is not acidic. A solution of NaHCO3 has a pH of very close to 8.3 and that is basic although not highly basic.
NaHCO3 + CH3COOH ==> CH3COONa + H2CO3, then the H2CO3, being unstable, decomposes to H2O and CO2 so you write it all at once to get
NaHCO3 + CH3COOH ==> H2O + CO2 + CH3COONa
To determine the third product of the reaction between CH3COOH (acetic acid) and NaHCO3 (sodium bicarbonate), we need to consider their chemical properties and reactiveness.
When acetic acid (CH3COOH) reacts with sodium bicarbonate (NaHCO3), the following two products are formed:
1) Sodium acetate (CH3COONa): This is formed by the combination of the acetate ion (CH3COO-) from acetic acid and the sodium ion (Na+) from sodium bicarbonate. Sodium acetate is a weak acid salt.
2) Carbon dioxide (CO2) gas: This is formed when sodium bicarbonate decomposes due to the reaction with the acidic acetic acid. The reaction releases carbon dioxide gas (CO2), water (H2O), and sodium acetate (CH3COONa).
So, the first two products you mentioned are correct: CH3COO^-(Na)^+ and CO2(g).
Regarding the third product, it would not be Na2CO3. Sodium carbonate (Na2CO3) is not formed in this reaction. Instead, the third product is water (H2O).
Therefore, the complete reaction can be written as:
CH3COOH + NaHCO3 → CH3COO^-(Na)^+ + CO2(g) + H2O
It's important to note that sodium acetate (CH3COONa) itself is not a weak acid. It is a weak acid salt because it is the conjugate base of the weak acid, acetic acid (CH3COOH).
Please let me know if I can help you with anything else.