Question

A 500 ml of $0.1$N solution of $\text{AgN}{{\text{O}}_{\text{3}}}$ is added to 500 mL of $0.1$ N solution of $\text{KCl}$. The concentration of the nitrate ion in the resulting mixture is:
A. $0.05$N
B. $0.1$N
C. $0.2$N
D. Reduced to zero

Answer 1

Hint: The normality of a solution can be defined as the no. of gram equivalents of the substance dissolved in one litre of the solution. It can be calculated from the formula given. We shall calculate the normality of the nitrate ion by substituting the values in the equation.

Formula used:
 Mathematically,
Gram equivalent weight = $\dfrac{\text{M}\text{.W}\text{.}}{\text{n}}$

Complete step by step answer:
Firstly we need to find the normality of 500 mL of $0.1$N solution of $\text{KCl}$= $\dfrac{500}{1000}\times 0.1$= $0.05$ N.
Similarly the normality of 500 ml of $0.1$N solution of $\text{AgN}{{\text{O}}_{\text{3}}}$= $0.05$ N
After the two solutions react with each other, they form potassium nitrate and the precipitate of silver chloride.
The total volume of the solution is now 1 litre, from the 500 mL of both the solutions. Hence the normality of the solution = number of gram equivalents of the solute per litre of the solution.
As the number of gram equivalents of silver nitrate = $0.05$ N
Hence, the same number will be there in the final solution too.

So, the correct answer is option A.

Note:
Gram equivalent weight of a substance is defined as the mass in grams that is numerically equal to the equivalent weight of the substance which is the molecular weight divided by the number of moles of the substance.
Molarity is another such unit of concentration which is defined as the number of moles of a solute present in one litre of the solution.