
Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid \[\left( {HCl} \right)\] is added to test-tube A, while acetic acid \[\left( {C{H_3}COOH} \right)\] is added to test tube B. In which test tube will have fizzing occur more vigorously?
A. A
B. B
C. Both
D. None
Answer
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Hint: As we know that fizzing occurs in a chemical reaction due to the presence of a gas. Here, to identify the test tube in which fizzing occurs more vigorously, then we need to compare the acidity of both \[C{H_3}COOH\] and \[HCl\].
Complete step by step answer:
According to question when equal lengths of magnesium ribbons are taken in test tubes A and B. Then, hydrochloric acid \[\left( {HCl} \right)\] is added to test-tube A, while acetic acid \[\left( {C{H_3}COOH} \right)\] is added to test tube B.
Thus, the reaction in test tube A is:
$Mg(s) + 2HCl(aq) \to MgC{l_2}(aq) + {H_2}(g)$
And the reaction in test tube B is:
$Mg(s) + C{H_3}COOH(aq) \to Mg{(C{H_3}COO)_2}(aq) + {H_2}(g)$
It is evident that hydrogen gas is released in both the reactions. Though concentration of both the acids is same but we also know that hydrochloric acid is a stronger and reactive acid than acetic acid and therefore, the hydrogen gas produced in the reaction is at a faster rate due to which frizzing occurs in test tube A.
As \[HCl\] is a stronger acid as compared to$C{H_3}COOH$, magnesium metal is added to the chemical reaction. Thus, \[HCl\] produces hydrogen gas due to which fizzing will be more vigorous in test-tube A.
So, the correct answer is Option A .
Note:
From the above question, we get to know that fizzing will occur more in test tube A because:
\[HCl\] is a stronger acid whereas $C{H_3}COOH$ is a weaker acid.
Hydrochloric acid completely dissociates into ${H^ + }$ and $C{l^ - }$ions while $C{H_3}COOH$doesn’t ionize completely.
Complete step by step answer:
According to question when equal lengths of magnesium ribbons are taken in test tubes A and B. Then, hydrochloric acid \[\left( {HCl} \right)\] is added to test-tube A, while acetic acid \[\left( {C{H_3}COOH} \right)\] is added to test tube B.
Thus, the reaction in test tube A is:
$Mg(s) + 2HCl(aq) \to MgC{l_2}(aq) + {H_2}(g)$
And the reaction in test tube B is:
$Mg(s) + C{H_3}COOH(aq) \to Mg{(C{H_3}COO)_2}(aq) + {H_2}(g)$
It is evident that hydrogen gas is released in both the reactions. Though concentration of both the acids is same but we also know that hydrochloric acid is a stronger and reactive acid than acetic acid and therefore, the hydrogen gas produced in the reaction is at a faster rate due to which frizzing occurs in test tube A.
As \[HCl\] is a stronger acid as compared to$C{H_3}COOH$, magnesium metal is added to the chemical reaction. Thus, \[HCl\] produces hydrogen gas due to which fizzing will be more vigorous in test-tube A.
So, the correct answer is Option A .
Note:
From the above question, we get to know that fizzing will occur more in test tube A because:
\[HCl\] is a stronger acid whereas $C{H_3}COOH$ is a weaker acid.
Hydrochloric acid completely dissociates into ${H^ + }$ and $C{l^ - }$ions while $C{H_3}COOH$doesn’t ionize completely.
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