Question

A mixture of ${{\text{H}}_{2}}{{\text{C}}_{2}}{{\text{O}}_{4}}$ and $\text{NaH}{{\text{C}}_{2}}{{\text{O}}_{4}}$ weighing 2.02 g was dissolved in water and the solution made up to one litre. 10 ml of this solution required 3.0 ml of 0.1 N NaOH solution for complete neutralization. In another experiment 10 ml of same solution in hot dilute ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ medium required 4 ml of 0.1 N $\text{KMn}{{\text{O}}_{4}}$ for complete neutralization. Calculate the amount of ${{\text{H}}_{2}}{{\text{C}}_{2}}{{\text{O}}_{4}}$and $\text{NaH}{{\text{C}}_{2}}{{\text{O}}_{4}}$ in the mixture.

Answer 1

Hint: Using the concept of normality i.e. the no of moles of gram equivalent of the solute to the total volume of the solution in 1L(1000ml), we can calculate the mass of solutes i.e. ${{\text{H}}_{2}}{{\text{C}}_{2}}{{\text{O}}_{4}}$and $\text{NaH}{{\text{C}}_{2}}{{\text{O}}_{4}}$.

Complete step by step answer:
The reactions between the acid and bases are known as the neutralization reactions and in this, salt and water are obtained as the products.
We know that the total mixture of ${{\text{H}}_{2}}{{\text{C}}_{2}}{{\text{O}}_{4}}$ and $\text{NaH}{{\text{C}}_{2}}{{\text{O}}_{4}}$ dissolved in the water is about 2.02(given). Let the amount of ${{\text{H}}_{2}}{{\text{C}}_{2}}{{\text{O}}_{4}}$ and $\text{NaH}{{\text{C}}_{2}}{{\text{O}}_{4}}$in the reaction mixture be as a and b. Then, a + b = 2.02  (1) For 1L mixture, the reaction occurs as:
${{\text{H}}_{2}}{{\text{C}}_{2}}{{\text{O}}_{4}}\text{+NaH}{{\text{C}}_{2}}{{\text{O}}_{4}}\text{ }\to \text{NaOH}$
As in the statement, the normality is given as:0.1, as we know that,
Normality=$\dfrac {\text {no of moles of gram equivalent of the solute}} {\begin{align} & \text {total volume of the solution in} \\ & \text{1L(1000ml)} \\
\end{align}} $
We know the normality and the total volume of the solvent ,then therefore the no of moles of the solute ${{\text{H}}_{2}}{{\text{C}}_{2}}{{\text{O}}_{4}}$ and $\text{NaH}{{\text{C}}_{2}}{{\text{O}}_{4}}$ can be calculated:
Normality × total volume of the solution in 1L = no of moles of the gram equivalent of solutes (2)
The no of moles of the solute can be calculated by the following as:
No of moles of the solute= $\dfrac{\text{given mass of}{{\text{H}}_{2}}{{\text{C}}_{2}}{{\text{O}}_{4}}\text{ }}{\text{equivalent molar mass}}\times 1000+\dfrac{\text{given mass of NaH}{{\text{C}}_{2}}{{\text{O}}_{4}}}{\text{ equivalent molar mass}}$ × 1000 (3)
Since, we are calculating the normality, therefore, in case of it, equivalent molar mass is taken which is obtained by dividing the molar mass by 2.
Molar mass of ${{\text{H}} _ {2}} {{\text{C}} _ {2}} {{\text{O}} _ {4}} $ = 2×24×64
                                      = 90
So, equivalent molar mass of ${{\text{H}} _ {2}} {{\text{C}} _ {2}} {{\text{O}} _ {4}} $ = $\dfrac {90}{2} $
                                                              = 45
Similarly, molar mass of $\text {NaH}{{\text{C}} _ {2}} {{\text{O}} _ {4}} $ = 23 × 1 × 24 × 64
                                                          = 112
So, the equivalent molar mass of $\text {NaH}{{\text{C}} _ {2}} {{\text{O}} _ {4}} $ = $\dfrac {112}{2} $
                                                                       = 56
Since, we assumed the molar mass of ${{\text{H}}_{2}}{{\text{C}}_{2}}{{\text{O}}_{4}}$ as ‘a’ and molar mass of $\text{NaH}{{\text{C}}_{2}}{{\text{O}}_{4}}$ as ‘b’, substituting all these value in eq (3), we get:
Number of moles of the solute = $\dfrac{\text{a}}{45} \times 1000+\dfrac{\text{b}} {\text {56}} $ × 1000
Now, put all the values in eq (2), we get:
 $\dfrac{\text{a}}{45} \times 1000+\dfrac{\text{b}} {\text {56}} $ × 1000 = 3 × 0.1 × $\dfrac {1000}{10} $ (4)
Similarly, for the redox changes i.e. during the second neutralization reaction of $\text {KMn}{{\text{O}} _ {4}} $ for 1L (1000ml) of mixture, the reaction occurs as:
 ${{\text{H}}_{2}}{{\text{C}}_{2}}{{\text{O}}_{4}}\text{+NaH}{{\text{C}}_{2}}{{\text{O}}_{4}}\text{ }\to \text{KMn}{{\text{O}}_{4}}$
And similarly, the normality for the reaction is given as:
$\dfrac{\text{a}}{45} \times 1000+\dfrac{\text{b}} {\text {56}} $×1000 = 4×0.1×$\dfrac {1000}{10} $ (5)
On Solving Equations, the equations (1) and (4) and (5), we get, a = 0.90 g
           b = 1.12 g
So, thus, mass of ${{\text{H}} _ {2}} {{\text{C}} _ {2}} {{\text{O}} _ {4}} $ = 0.90g
          mass of $\text {NaH}{{\text{C}} _ {2}} {{\text{O}} _ {4}} $ = 1.12g

Note: The normality of the solution changes with the temperature due to changes accompanied in the volume of the solution and thus, the normality of the solution always depends on the temperature.