When you add a few drops of acetic acid to a test-tube containing sodium bicarbonate powder, which one of the following is your observation?
(A)- No reaction takes place.
(B)- A colourless gas with pungent smell is released with brisk effervescence.
(C)- A brown coloured gas is released with brisk effervescence.
(D)- Formation of bubbles of a colourless and odourless gas.
Answer
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Hint: The acid and the base react to form the salt and water, along with the release of gas. This gas turns lime water milky and is used commonly in fire-extinguishers.
Complete step by step answer:
On adding the drops of acetic acid, $C{{H}_{3}}COOH$ in the test tube having sodium carbonate powder, $NaHC{{O}_{3}}$, the neutralization reaction occurs through the following steps as follows:
- Firstly, the ions (cation or anion) from both the reactants displace one another, leading to the double displacement reaction forming sodium acetate and carbonic acid as shown:
$NaHC{{O}_{3}}(aq)+C{{H}_{3}}COOH(aq)\to C{{H}_{3}}COONa(aq)+{{H}_{2}}C{{O}_{3}}(aq)$
The carbonic acid formed is highly unstable in nature and will now lead to the second step of the reaction.
- The carbonic acid decomposes to produce water and carbon-dioxide gas as shown:
${{H}_{2}}C{{O}_{3}}(aq)\,\,\rightleftarrows {{H}_{2}}O(l)+C{{O}_{2}}(g)\uparrow $
The carbon-dioxide gas released is colourless and odourless and shows the vigorous evolution of the gas as it escapes the solution producing brisk effervescence.
So, we get the overall reaction, which involves the formation of sodium acetate salt by the reaction of acid and base, shows the neutralization reaction as follows:
$C{{H}_{3}}COOH(aq)+NaHC{{O}_{3}}(aq)\to C{{H}_{3}}COONa(aq)+{{H}_{2}}O(l)+C{{O}_{2}}(g)\uparrow $
Therefore, in the reaction it is observed option (B)- A colourless gas with pungent smell is released with brisk effervescence, that is, the carbon-dioxide gas.
So, the correct answer is “Option B”.
Note: The presence of the gas can be tested by passing the evolved gas through lime water, which turns milky.
- This gas formed during the reaction, is used in fire extinguishers as it is denser than the air. On spraying it over the fire, it cuts off the combustion by the oxygen as it sinks under the oxygen layer and separates it. Thus, stops the fire from spreading.
Complete step by step answer:
On adding the drops of acetic acid, $C{{H}_{3}}COOH$ in the test tube having sodium carbonate powder, $NaHC{{O}_{3}}$, the neutralization reaction occurs through the following steps as follows:
- Firstly, the ions (cation or anion) from both the reactants displace one another, leading to the double displacement reaction forming sodium acetate and carbonic acid as shown:
$NaHC{{O}_{3}}(aq)+C{{H}_{3}}COOH(aq)\to C{{H}_{3}}COONa(aq)+{{H}_{2}}C{{O}_{3}}(aq)$
The carbonic acid formed is highly unstable in nature and will now lead to the second step of the reaction.
- The carbonic acid decomposes to produce water and carbon-dioxide gas as shown:
${{H}_{2}}C{{O}_{3}}(aq)\,\,\rightleftarrows {{H}_{2}}O(l)+C{{O}_{2}}(g)\uparrow $
The carbon-dioxide gas released is colourless and odourless and shows the vigorous evolution of the gas as it escapes the solution producing brisk effervescence.
So, we get the overall reaction, which involves the formation of sodium acetate salt by the reaction of acid and base, shows the neutralization reaction as follows:
$C{{H}_{3}}COOH(aq)+NaHC{{O}_{3}}(aq)\to C{{H}_{3}}COONa(aq)+{{H}_{2}}O(l)+C{{O}_{2}}(g)\uparrow $
Therefore, in the reaction it is observed option (B)- A colourless gas with pungent smell is released with brisk effervescence, that is, the carbon-dioxide gas.
So, the correct answer is “Option B”.
Note: The presence of the gas can be tested by passing the evolved gas through lime water, which turns milky.
- This gas formed during the reaction, is used in fire extinguishers as it is denser than the air. On spraying it over the fire, it cuts off the combustion by the oxygen as it sinks under the oxygen layer and separates it. Thus, stops the fire from spreading.
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