Equal volumes of 0.1 M AgNO_3 and 0.2 M NaC | vedclass

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(B) Let the volume of each solution be $V \ L$.Initial millimoles of $AgNO_3 = 0.1 \ M 	imes V \ L = 0.1V \ mmol$.Since $AgNO_3$ dissociates complete

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$0.05$

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(B) Let the volume of each solution be $V \ L$.Initial millimoles of $AgNO_3 = 0.1 \ M 	imes V \ L = 0.1V \ mmol$.Since $AgNO_3$ dissociates completely as $AgNO_3 ightarrow Ag^+ + NO_3^-$,the millimoles of $NO_3^-$ ions are equal to the millimoles of $AgNO_3$,which is $0.1V \ mmol$.The total volume of the mixture $= V + V = 2V \ L$.The concentration of $NO_3^-$ in the final mixture is given by:$[NO_3^-] = \frac{	ext{Total millimoles of } NO_3^-}{	ext{Total volume}} = \frac{0.1V}{2V} = 0.05 \ M$.

Equal volumes of $0.1 \ M$ $AgNO_3$ and $0.2 \ M$ $NaCl$ are mixed. The concentration of $NO_3^-$ ion in the mixture will be ............ $M$.

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