Question

Volume \[40mL\] of \[0.05M{ }\]\[N{{a}_{2}}C{{O}_{3}}.NaHC{{O}_{3}}.2{{H}_{2}}O\] (sesquicarbonate) is titrated against $0.05M$$HCl$. \[xmL\] of $HCl$ is used when phenolphthalein is indicated and $ymL$ of $HCl$ is used when methyl orange is the indicator in two separate titrations hence $(y-x)$ is --
A. $80mL$
B. $30mL$
C. $120mL$
D. none of the above

Answer 1

Hint: The difference in volume of hydrochloric acid which was used in the process of titration with two different indicators which are phenolphthalein and methyl orange, can be calculated by simply using the given volume of the titrant which is sesquicarbonate in this case.
The concepts of molarity and normality are being used in order to calculate the individual volumes of the compounds.

Complete step-by-step answer: In order to solve this question, we must first write down all the given values of the quantities to get a better idea of the question itself. The molarity of the sesquicarbonate which has the formula \[N{{a}_{2}}C{{O}_{3}}.NaHC{{O}_{3}}.2{{H}_{2}}O\] is given as \[0.05M{ }\]and the volume of the sesquicarbonate is indicated as \[40mL\]. The volume of hydrochloric acid which has the formula $HCl$ is \[xmL\] when we are using phenolphthalein as an indicator and its $ymL$ when we are using methyl orange as an indicator. And also the concentration of this $HCl$ \[0.05M{ }\].
Now we will calculate the normality of the solutions by using a unitary method. We know that \[0.05M{ }\] of sodium carbonate containing solution has the normality of $0.1N$. This value calculation is done by using the relationship between molarity and normality. Normality is equal to ‘n’ times the molarity of the same substance, where ‘n’ stands for the acidity or the basicity of the substance. We know in case of sodium carbonate the value of ‘n’ is two, because each molecule of sodium carbonate would give two ions of sodium. Hence, we get the normality as $0.1N$.
Now, we would calculate the normality of sodium bicarbonate in the same way, we get,
\[0.05~M~NaHC{{O}_{3}}=0.05~N~NaHC{{O}_{3}}\]
Now, we would relate the volume and molarity of sodium carbonate and hydrochloric acid, we get,
\[40~mL~\] of \[0.1~N{{a}_{2}}C{{O}_{3}}\]\[=40~mL\] of \[0.1~N~HCl\]
And the volume of hydrochloric acid is determined by using the formula ${{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}$ where ${{N}_{1}},{{V}_{1}}$ are the normality and volume of sodium carbonate and the ${{N}_{2}},{{V}_{2}}$ are the normality and volume of the hydrochloric acid respectively. For the complete reaction $80mL$ of hydrochloric acid is required as half of which would be used with phenolphthalein indicator and the other half would be used with methyl orange indicator. So, we assume $x=40mL$.
With phenolphthalein
$40mL$ of $0.05MNaHC{{O}_{3}}$ $=40mL$ of $0.05MHCl$
And with methyl orange $y=80+40=120mL$
So, now we will calculate the value of $(y-x)$ which is $120-40=80mL$.

Clearly the correct answer would be option A which is $80mL$.

Note: The molarity of a substance is the number of moles of that substance or the solute present in one litre of the solution.
The relationship between molarity and normality is that, the molarity is equal to the normality of the acidity or the basicity of a compound is unity.