Equal volume of 0.1 M $A g N O_{3}$ and 0.2 M NaCl are mixed. The concentration of $N O_{3}^{-}$ ions in the mixture will be:

Open in App

Solution

Assuming that the concentration of NaCl is also 0.1 M
Suppose volumes mixed are V mL then m mole $A g N O_{3} = N O_{3} = 0.1 \times V$
Now after mixing total vol. = 2 V
So, concentration of $N O_{3}$ ions (which are spectators in reaction) after mixing = $0.1 \times \frac{V}{2 V} = 0.05 M$

Suggest Corrections

Similar questions

0.1 M A g N O_{3}

0.2 M N a C l

N O_{3}^{-}

Equal volumes of 0.1M $A g N O_{3}$ and 0.2 M NaCl are mixed together. The concentration of nitrate ions in the resulting mixture is:

0.1 M

A g {N O}_{3}

0.2 M

N a C l

{N O}_{3}^{-}

A g N O_{3}

N O_{3}^{-}

0.1 M A g N O_{3}

0.1 M N a C l

Join BYJU'S Learning Program