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Stoichiometric Calculations

500 ml of a 0...

Question

500 ml of a 0.1 N solution of $A g {N O}_{3}$ is added to 500 ml of 0.1 N solution of KCl. The concentration of nitrate ion in the resulting mixture is
(a) 0.05 N (b) 0.1 N (c) 0.2 N (d) reduced to zero

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Solution

The correct answer is a gram equivalent of $N o_{3}$ = $0.500 \times 0.1$
there are no gmeqs of $N o_{3}$ in $K C l$
total volume= $500 m l + 500 m l$ =1L
hence concentration of $N o_{3}$ = $\frac{0.05}{1}$
=0.05 N

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