
Assertion: Carbonates and bicarbonates are oxidized to \[C{O_2}\], upon addition of strong acid in it.
Reason: \[C{O_2}\] is the only gas which turns lime water milky
A. Both Assertion and Reason are correct and reason is the correct explanation for Assertion.
B. Both assertion and reason are correct but reason is not the correct explanation for Assertion.
C. Assertion is correct but reason is incorrect.
D. Both assertion and reason is incorrect.
Answer
589.2k+ views
Hint: We must understand that the lime water turns milky due to the formation of insoluble suspension in the solution. Also, when bicarbonates \[({HCO_3}^- )\] are oxidized they produce \[{H_2}O\].
Complete step to step solution:
Let’s check out the assertion first. According to the assertion, ‘Carbonates and bicarbonates are oxidized to \[C{O_2}\], upon addition of strong acid in it.’ We know that the carbonates and the bicarbonates both when heated are oxidized to \[C{O_2}\].
Let’s take an example of heating of Sodium bicarbonate
\[2{\text{NaHC}}{{\text{O}}_3}\xrightarrow{\Delta }{\text{ N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}{\text{ + }}{{\text{H}}_2}{\text{O + C}}{{\text{O}}_2}\]
Similarly in the presence of weak acid also they are oxidized to $CO_2$.
Let’s take an example of Sodium bicarbonate reacting with acetic acid which is a weak acid.
\[{\text{NaHC}}{{\text{O}}_{\text{3}}}{\text{ + C}}{{\text{H}}_{\text{3}}}{\text{COOH}} \to {\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}}{\text{ + }}{{\text{H}}_2}{\text{O}}{\text{ + C}}{{\text{O}}_{{\text{2}}}}\]
But when a strong acid reacts with a carbonate it forms a salt, along with hydrogen carbonate.
For this let’s take an example of Sodium carbonate reacting with strong \[HCl\]
${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}{\text{ + HCl }} \to {\text{ }}{{\text{H}}_2}{\text{C}}{{\text{O}}_3}{\text{ + NaCl}}$
This means that the assertion is wrong.
Let’s move towards the reason, we all know that the \[C{O_2}\] turns the lime water milky. But is it the only gas which turns the lime water milky. Well, the answer is no.
$SO_2$ also turns the lime water milky. Here’s the reaction
\[{\text{Ca(OH}}{{\text{)}}_{\text{2}}}{\text{ }}{\text{ + S}}{{\text{O}}_{\text{2}}}{\text{ }} \to {\text{ CaS}}{{\text{O}}_{\text{4}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\]
Milky colour is formed due to the presence of \[CaS{O_4}\].
So the answer to this question is D. both assertion and reason are incorrect.
Note: Sodium bicarbonates are a salt of sodium which breaks to form sodium and bicarbonate in water. This breakdown makes it an alkaline solution which means it can neutralize acid. That is why sodium bicarbonate is often used to treat conditions caused by high body acidity, such as heartburn.
Carbonate \[\left( {N{a_2}C{O_3}} \right)\] is a sodium salt of carbonic acid and a relatively strong, non-volatile base. We also know it as soda washing. Crystalline heptahydrate is the most commonly occurring salt form. That’s efflorescence readily to form a white powder, the monohydrate. It has an alkaline fresh taste, and can be extracted from many plants' ashes. We can artificially produce this from common salt in large quantities.
And also while answering such questions we must remember that there is a possibility of assertion being correct and the reason being wrong, so students need to check both the assertion and the reason for answering the question correctly.
Complete step to step solution:
Let’s check out the assertion first. According to the assertion, ‘Carbonates and bicarbonates are oxidized to \[C{O_2}\], upon addition of strong acid in it.’ We know that the carbonates and the bicarbonates both when heated are oxidized to \[C{O_2}\].
Let’s take an example of heating of Sodium bicarbonate
\[2{\text{NaHC}}{{\text{O}}_3}\xrightarrow{\Delta }{\text{ N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}{\text{ + }}{{\text{H}}_2}{\text{O + C}}{{\text{O}}_2}\]
Similarly in the presence of weak acid also they are oxidized to $CO_2$.
Let’s take an example of Sodium bicarbonate reacting with acetic acid which is a weak acid.
\[{\text{NaHC}}{{\text{O}}_{\text{3}}}{\text{ + C}}{{\text{H}}_{\text{3}}}{\text{COOH}} \to {\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}}{\text{ + }}{{\text{H}}_2}{\text{O}}{\text{ + C}}{{\text{O}}_{{\text{2}}}}\]
But when a strong acid reacts with a carbonate it forms a salt, along with hydrogen carbonate.
For this let’s take an example of Sodium carbonate reacting with strong \[HCl\]
${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}{\text{ + HCl }} \to {\text{ }}{{\text{H}}_2}{\text{C}}{{\text{O}}_3}{\text{ + NaCl}}$
This means that the assertion is wrong.
Let’s move towards the reason, we all know that the \[C{O_2}\] turns the lime water milky. But is it the only gas which turns the lime water milky. Well, the answer is no.
$SO_2$ also turns the lime water milky. Here’s the reaction
\[{\text{Ca(OH}}{{\text{)}}_{\text{2}}}{\text{ }}{\text{ + S}}{{\text{O}}_{\text{2}}}{\text{ }} \to {\text{ CaS}}{{\text{O}}_{\text{4}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\]
Milky colour is formed due to the presence of \[CaS{O_4}\].
So the answer to this question is D. both assertion and reason are incorrect.
Note: Sodium bicarbonates are a salt of sodium which breaks to form sodium and bicarbonate in water. This breakdown makes it an alkaline solution which means it can neutralize acid. That is why sodium bicarbonate is often used to treat conditions caused by high body acidity, such as heartburn.
Carbonate \[\left( {N{a_2}C{O_3}} \right)\] is a sodium salt of carbonic acid and a relatively strong, non-volatile base. We also know it as soda washing. Crystalline heptahydrate is the most commonly occurring salt form. That’s efflorescence readily to form a white powder, the monohydrate. It has an alkaline fresh taste, and can be extracted from many plants' ashes. We can artificially produce this from common salt in large quantities.
And also while answering such questions we must remember that there is a possibility of assertion being correct and the reason being wrong, so students need to check both the assertion and the reason for answering the question correctly.
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