Question

One ${\text{g}}$ of a mixture of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ and ${\text{NaHC}}{{\text{O}}_{\text{3}}}$ consumes ${\text{y}}$ equivalent ${\text{HCl}}$ for complete neutralization. One ${\text{g}}$ of the mixture is strongly heated, then cooled and the residue treated with ${\text{HCl}}$. How many equivalents of ${\text{HCl}}$ would be required for complete neutralization?
(A)- ${\text{2y}}$ equivalent
(B)- ${\text{y}}$ equivalent
(C)- $\frac{{{\text{3y}}}}{{\text{4}}}$ equivalent
(D)- $\frac{{{\text{3y}}}}{2}$ equivalent

Answer 1

Hint: Neutralization is a process where we get a neutral solution of any substance by adding acid in a basic solution or by adding base in an acidic solution to make them neutral. pH of neutralized substance or solution is equal to 7 or near about 7.

Complete answer: Given that, for the complete neutralization of One ${\text{g}}$ of a mixture of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ and ${\text{NaHC}}{{\text{O}}_{\text{3}}}$ ;${\text{y}}$ equivalent ${\text{HCl}}$ is required.
Now they asked how much of ${\text{HCl}}$ is required for neutralizing the residue after the heating and cooling process.
As we know that on heating, sodium bicarbonate (${\text{NaHC}}{{\text{O}}_{\text{3}}}$) converts into sodium carbonate (${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$) which doesn’t convert to any element on further heating. It means amount of formed ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ is equals to the amount of heated ${\text{NaHC}}{{\text{O}}_{\text{3}}}$.
${\text{y}}$ equivalent of ${\text{HCl}}$ is required for neutralization when mixture contains half amount of ${\text{NaHC}}{{\text{O}}_{\text{3}}}$ and ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$. After heating formed residue is only ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ because all ${\text{NaHC}}{{\text{O}}_{\text{3}}}$ convert into ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ by following process:
${\text{2NaHC}}{{\text{O}}_{\text{3}}}\xrightarrow{{\text{\Delta }}}{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O + C}}{{\text{O}}_{\text{2}}}$.
So, amount of ${\text{HCl}}$ required for the complete neutralization of the residue after heating & cooling is equal to the ${\text{y}}$ i.e. option (b), which is equal to the amount of ${\text{HCl}}$ used for neutralization of One ${\text{g}}$ of a mixture of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ and ${\text{NaHC}}{{\text{O}}_{\text{3}}}$.

Note: Here you may get confuse in the calculation part for calculating amount of ${\text{HCl}}$ before and after the heating process, but always keep in mind when they are asking for the amount of required ${\text{HCl}}$ for neutralization of residue after heating process is equal to the amount of ${\text{HCl}}$ used for the mixture previously. If they are asking for the total amount of required ${\text{HCl}}$ after heating the mixture so it is equal to ${\text{2y}}$.