On adding $\text{NaHCO}_3$ to acetic acid, a gas is evolved which turns lime water milky due to the formation of:

<b>Correct Option (C):</b> When acetic acid ($\text{CH}_3\text{COOH}$) reacts with sodium hydrogencarbonate ($\text{NaHCO}_3$), carbon dioxide gas ($\text{CO}_2$) is evolved along with the formation of sodium acetate and water:<br>$$\text{CH}_3\text{COOH} + \text{NaHCO}_3 \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2 \uparrow$$<br>When the evolved carbon dioxide gas is passed through lime water (calcium hydroxide, $\text{Ca(OH)}_2$), it reacts to form an insoluble white precipitate of calcium carbonate ($\text{CaCO}_3$), which turns the solution milky:<br>$$\text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3 \downarrow + \text{H}_2\text{O}$$<br><br><b>Incorrect Option (A):</b> Calcium bicarbonate ($\text{Ca(HCO}_3)_2$) is highly soluble in water. It is formed only when excess carbon dioxide is passed through the milky solution, which makes the solution clear again.<br><b>Incorrect Option (B):</b> Calcium hydroxide is the chemical name for lime water itself, which is the reactant, not the product causing milkiness.<br><b>Incorrect Option (D):</b> Calcium acetate is a highly soluble salt and does not form a precipitate to turn the solution milky.