Equal volumes of $0.1 M$ $A g {N O}_{3}$ and $0.2 M$ $N a C l$ are mixed. The concentration of ${N O}_{3}^{-}$ ions in the mixture will be:

0.1 M

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0.05 M

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0.2 M

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0.15 M

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Solution

The correct option is B $0.05 M$

$[N O_{3}^{-}] i n 0.1 M A g N O_{3} = 0.1 M$
when mixed with an equal volume of $0.2 M$ $N a C l$ solution, the volume of the solution is double.
So the concentration of $[N O_{3}^{-}]$ is halved
$[N O_{3}^{-}]$
$= 0.1 / 2$
$= 0.05 M$
Hence, $0.05 M$ is the answer.

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