Question

14.3 g of $N{{a}_{2}}C{{O}_{3}}.x{{H}_{2}}O$ completely neutralizes 100 ml of 1N ${{H}_{2}}S{{O}_{4}}$ solutions, value of x is:

Answer 1

Hint: In this, both the acid and base undergoes complete neutralization, so, their gram equivalents will be equal to each other and we can easily calculate their gram equivalents by applying the formulas as: $\dfrac{given\text{ }mass}{equivalent\text{ }weight}$( to calculate gram equivalent for $N{{a}_{2}}C{{O}_{3}}.x{{H}_{2}}O$) and $normality\times given\text{ }volume\text{ }in\text{ }liters$ ( to calculate gram equivalent for ${{H}_{2}}S{{O}_{4}}$). Now solve it.

Complete Solution :
First let's discuss what neutralization reactions are. By the neutralization reactions, we mean the reactions in which the acid and base neutralizes each other and results in the formation of salt and water.
Now considering the statement:
As in the question, it is given that: $N{{a}_{2}}C{{O}_{3}}.x{{H}_{2}}O$ completely neutralizes ${{H}_{2}}S{{O}_{4}}$, then:
No. of gram equivalent of the $N{{a}_{2}}C{{O}_{3}}.x{{H}_{2}}O$ = no of gram equivalent of ${{H}_{2}}S{{O}_{4}}$----(1)
So, the gram equivalent of $N{{a}_{2}}C{{O}_{3}}.x{{H}_{2}}O$$ = \dfrac{given\text{ }mass}{equivalent\text{ }weight}$
Given mass=14.3 g (given)
Equivalent weight $ = \dfrac{molecular\text{ }mass}{2}$
As we know,
\[\begin{align}
  & Molar\text{ }mass\text{ }of\text{ }N{{a}_{2}}C{{O}_{3}}.x{{H}_{2}}O=23 \times 2+12+16\times 3+x(2+16) \\
 & \text{ = 46+12+48+x(18)} \\
 & \text{ = 106+18 x} \\
\end{align}\]

Then, the equivalent weight of $N{{a}_{2}}C{{O}_{3}}.x{{H}_{2}}O$$\text{ =}\dfrac{\dfrac{14.3}{106+18x}}{2}$ ----------(2)
The no. of gram equivalent of ${{H}_{2}}S{{O}_{4}}$$=normality\times given\text{ }volume\text{ }in\text{ }liters$
As, we know that; normality = 1N (given)
And volume= $100\text{ }ml = 100\times {{10}^{-3}}L$
Then
The no. of gram equivalent of ${{H}_{2}}S{{O}_{4}}$$\begin{align}
 & =normality\times given\text{ }volume\text{ }in\text{ }liters \\
 & =1\times 100\times {{10}^{-3}} \\
\end{align}$ -------(3)

Now, put the value of equations (2) and (3) in equation (1), we get;
$\begin{align}
 & \text{ }\dfrac{\dfrac{14.3}{106+18x}}{2}=1\times 100\times {{10}^{-3}} \\
 & 14.3\times 2=0.1(106+18x) \\
 & 28.6=10.6+1.8x \\
 & 18=1.8x \\
 & x=10 \\
\end{align}$

Hence, when 14.3 g of $N{{a}_{2}}C{{O}_{3}}.x{{H}_{2}}O$ completely neutralizes 100 ml of 1N ${{H}_{2}}S{{O}_{4}}$ solutions ,value of x is 10.

Note: While calculating the gram equivalents always keep in mind that the volume should be in its standard units i.e. in litres and in case of neutralization reactions, gram equivalents of both the acid and base are always equal to each other.