Question

On reacting $NaHC{O_3}$ and acetic acid, the gas evolved turns ________.

Answer 1

Neutralization reaction: It is a chemical reaction in which an acid reacts with a base to form salt and water. The value of resultant $pH$ varies with the change in strength of an acid or a base. In case of reaction of a weak acid with a weak base, the pH value depends on the respective${K_a}$ and ${K_b}$ values.

Complete answer:
Acetic acid: The molecular formula of the acetic acid is $C{H_3}COOH$. The ${K_a}$value of acetic acid is $1.7 \times {10^{ - 5}}$ and we know that lower the ${K_a}$ value, lesser will be the strength of the acid. So, we can conclude that acetic acid is a weak acid.
Sodium Bicarbonate: Its molecular formula is $NaHC{O_3}$. The${K_b}$ value of sodium bicarbonate is $2.4 \times {10^{ - 8}}$ and we know that lower the ${K_b}$ value, lesser will be the strength of the base. So, we can say that sodium bicarbonate is a weak base.
Therefore, when $NaHC{O_3}$reacts with acetic acid, a neutralization reaction takes place and an aqueous salt of sodium acetate is formed along with the evolution of carbon dioxide gas.
The reaction involved in the process proceeds as follows:
$C{H_3}COOH + NaHC{O_3} \to C{H_3}COONa + {H_2}O + C{O_2} \uparrow$
The gas evolved in the reaction is carbon dioxide, which is a colourless and odourless gas. It appears as a brisk effervescence, so after the reaction bubbles in the solution can be observed.
Hence, on reacting $NaHC{O_3}$ and acetic acid i.e., $C{H_3}COOH$, the gas evolved turns lime water milky.

Note:
The presence of carbon dioxide can also be tested by passing it through lime water. When $C{O_2}$ gas is passed through the lime water, the formation of solid calcium carbonate takes place due to which the solution turns milky.