Equal volumes of $0.1 M$ $A g N O_{3}$ and $0.2 M$ $N a C l$ are mixed. The concentration of $N O_{3}^{-}$ ion 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 C $0.05 M$
When equal volume of two solutions are mixed, volume gets doubled and thus, concentration changes. The new concentration of nitrate ions in the mixture will be $\frac{M}{2} = \frac{0.1}{2} = 0.05 M$

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