Question

the concentration of NO3 ions when equal volumes of 0.1 M AgNO3 and 0.1 M Nacl are mixed together is ?

Answer 1

When both are mixed the concentration of NO3 will be reduced to half and it becomes 0.05M.

Answer 2

After mixing, the volume will be doubled, thus the $NO_3$ concentration will be reduced to half.

Explanation:

We know that 0.1 M of $AgNO_3$ contains 0.1 moles of nitrate ions, i.e., $\mathrm{NO}_{3}^{-}$ ions and 0.1 moles of $\mathrm{Ag}^{+}$ in 1 L of the solution.

It is given in the question that, it is mixed with 0.1 M of NaCl in equal volume. The chemical equation will be as such.

$\begin{array}{l}{\mathrm{AgNO}_{3}+\mathrm{NaCl} \rightarrow \mathrm{NaNO}_{3}+\mathrm{AgCl}} \\ {0.1 \mathrm{M} \ \ \ \ \quad 0.1 \mathrm{M} \ \quad \ 0.1 \mathrm{M} \quad \ \ \ 0.1 \mathrm{M}}\end{array}$

We know that,

$\mathrm{M}_{1} \mathrm{V}_{1}+\mathrm{M}_{2} \mathrm{V}_{2}+\ldots . .=\mathrm{M}_{\mathrm{f}} \mathrm{V}_{\mathrm{f}}$

$0.1 \times 100=\mathrm{M}_{\mathrm{f}} \times 100$

$\mathrm{M}_{\mathrm{f}}=0.1 \mathrm{M}$

Therefore,  

$\Rightarrow \frac{0.1}{2}=0.05 \ \mathrm{M}$