Question

1.2 g sample of $N{{a}_{2}}C{{O}_{3}}$and ${{K}_{2}}C{{O}_{3}}$was dissolved in water to form 100 mL of solution. 20 mL of this solution required 40 mL of 0.1 N HCl for complete neutralization. If another 20 mL of this solution is treated with excess $BaC{{l}_{2}}$, the weight of the precipitate is expressed as $\dfrac{197}{1000}x\text{ g}$then x is :

Answer 1

If the volume and normality of reactants of neutralization reactions are given, and we are to find the one unknown quantity of the reactants, we can use the following formula,
${{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}$

Complete step by step solution:
-Given, the total mass of mixture = 1.2 g
Volume of solution = 100 mL
Volume of solution taken for neutralization, ${{V}_{2}}$= 20 mL
Volume of HCl required for neutralizing, ${{V}_{1}}$ = 40 mL
Normality of HCl, ${{N}_{1}}$ = 0.1 N

-For finding the mass of $N{{a}_{2}}C{{O}_{3}}$in mixture, we need to start by calculating the normality of the solution,
${{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\Rightarrow 0.1\times 40={{N}_{2}}\times 20$
$\therefore {{N}_{2}}=0.2N$

-Now finding the gram equivalents of the solution,
Number of gram equivalents contained in 1 liter of solution = 0.2 gram equivalents
$\therefore$ Number of gram equivalents contained in 100 ml of solution $=0.2\times 0.1=0.02$ gram equivalents

-We know that the number of gram equivalents of the solution is equal to the summation of gram-equivalents of $N{{a}_{2}}C{{O}_{3}}$and ${{K}_{2}}C{{O}_{3}}$.
Number of gram equivalents $=\dfrac{mass}{\text{equivalent mass}}$
Equivalent mass $=\dfrac{\text{molar mass}}{\text{valence electrons}}$
where valence electrons are the number of electrons gained or lost by one molecule or ion of the substance in the reaction.

-Molar mass of $N{{a}_{2}}C{{O}_{3}}=22.989\times 2+12.010+15.999\times 3=105.987g\approx 106g$
Molar mass of ${{K}_{2}}C{{O}_{3}}=(39.098\times 2)+12.010+(15.999\times 3)=138.2055g\approx 138g$

-Equivalent mass of $N{{a}_{2}}C{{O}_{3}}=\dfrac{106}{2}=53g$
Equivalent mass of ${{K}_{2}}C{{O}_{3}}=\dfrac{138}{2}=69g$

-Let us assume the mass of $N{{a}_{2}}C{{O}_{3}}$= x g
Then the mass of ${{K}_{2}}C{{O}_{3}}$= (1.2-x) g

-So, we can write,
$0.02=\dfrac{(\dfrac{x}{53})+(1.2-x)}{69}=\dfrac{69x+63.6-53x}{3657}$
$\Rightarrow x=0.59\approx 0.6g$

-Further, the solution of$N{{a}_{2}}C{{O}_{3}}$and ${{K}_{2}}C{{O}_{3}}$is further treated with barium chloride to produce barium carbonate.
$\because$ 100 mL of solution contains = 0.02 g of equivalents
$\therefore$ 20 mL of solution will contain $=\dfrac{0.2}{100}\times 20$ = 0.004 g of equivalents

-Mass of Barium carbonate formed will be equivalent mass $\times$ number of gram-equivalents
Equivalent mass of barium carbonate $=\dfrac{137.2}{2}=68.65g$
Mass of barium carbonate $=0.004\times 68.65=0.2746g$

Note: The number of gram or mole equivalents of solute present in one litre of solution is known as the normality of the solution. The number of moles of reactive units present in a compound is known as the equivalents of the compound.