
In the mixture of \[NaHC{O_3}\] and \[N{a_2}C{O_3}\], volume of a given $HCl$ required is x ml with phenolphthalein indicator and further y ml is required with methyl orange indicator. Hence, volume of $HCl$ for complete reactions of \[NaHC{O_3}\] present in the original mixture is:
A.\[2x\]
B.$Y$
C.\[x/2\]
D.\[\left( {y - x} \right)\]
Answer
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Hint: We have to know that the sodium carbonate is a chemical compound having the molecular formula, \[N{a_2}C{O_3}\] and it has various hydrates. The sodium carbonate is white in colour and it does not have any odor. And the sodium carbonate is completely soluble in water and there is a formation of moderately basic solution. The sodium bicarbonate is a chemical compound having the molecular formula, \[NaHC{O_3}\] and it is also known as bicarbonate of soda or baking soda. And it contains sodium cation and bicarbonate anion.
Complete answer:
The volume of $HCl$ for complete reactions of \[NaHC{O_3}\] present in the original mixture is not equal to \[2x\]. Hence, option (A) is incorrect.
The volume of $HCl$ for the complete reactions of sodium bicarbonate is not equal to y. Hence, option (B) is incorrect.
The volume of hydrochloric acid for sodium bicarbonate is not equal to $\dfrac{x}{2}$. Hence, option (C) is incorrect.
The methyl orange specifies the complete neutralization of sodium carbonate and sodium bicarbonate. But in the case of phenolphthalein, it designates only the\[50\% \] reaction of sodium carbonate to sodium bicarbonate.
Let, volume of hydrochloric acid required to complete neutralization of \[NaHC{O_3} = {V_1}\]
And for the neutralization of sodium carbonate \[ = {V_2}\]
For phenolphthalein indicator \[\left[ {50\% N{a_2}C{O_3}} \right] = \dfrac{{{V_2}}}{2} = n\]
And for methyl orange \[ = \dfrac{{{V_2}}}{2} + {V_1} = y\]
Therefore, volume of hydrochloric acid for sodium bicarbonate, \[{V_1} = \left( {y - x} \right)ml\]
Hence, option (D) is correct.
Note:
We need to know that the indicator is a substance that shows colour when it is contacted with acidic and basic medium due to change in the pH of solution. Hence, by adding indicators, we can determine the pH of the solution visually. Phenolphthalein and methyl orange are examples of indicators. When the phenolphthalein is added to an acidic solution, it becomes colorless and if it is added in the basic solution, the solution becomes pink in colour. And the methyl orange in acid becomes red and in base, it becomes yellow.
Complete answer:
The volume of $HCl$ for complete reactions of \[NaHC{O_3}\] present in the original mixture is not equal to \[2x\]. Hence, option (A) is incorrect.
The volume of $HCl$ for the complete reactions of sodium bicarbonate is not equal to y. Hence, option (B) is incorrect.
The volume of hydrochloric acid for sodium bicarbonate is not equal to $\dfrac{x}{2}$. Hence, option (C) is incorrect.
The methyl orange specifies the complete neutralization of sodium carbonate and sodium bicarbonate. But in the case of phenolphthalein, it designates only the\[50\% \] reaction of sodium carbonate to sodium bicarbonate.
Let, volume of hydrochloric acid required to complete neutralization of \[NaHC{O_3} = {V_1}\]
And for the neutralization of sodium carbonate \[ = {V_2}\]
For phenolphthalein indicator \[\left[ {50\% N{a_2}C{O_3}} \right] = \dfrac{{{V_2}}}{2} = n\]
And for methyl orange \[ = \dfrac{{{V_2}}}{2} + {V_1} = y\]
Therefore, volume of hydrochloric acid for sodium bicarbonate, \[{V_1} = \left( {y - x} \right)ml\]
Hence, option (D) is correct.
Note:
We need to know that the indicator is a substance that shows colour when it is contacted with acidic and basic medium due to change in the pH of solution. Hence, by adding indicators, we can determine the pH of the solution visually. Phenolphthalein and methyl orange are examples of indicators. When the phenolphthalein is added to an acidic solution, it becomes colorless and if it is added in the basic solution, the solution becomes pink in colour. And the methyl orange in acid becomes red and in base, it becomes yellow.
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