When equal volumes of 0.1 M AgNO_3 and 0.2 | vedclass

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(B) Let the volume of each solution be $V \ L$. Total volume of the mixture = $V + V = 2V \ L$. The number of moles of $NO_3^-$ ions comes from $AgNO_

Vedclass · Accepted Answer

$0.05$

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(B) Let the volume of each solution be $V \ L$. Total volume of the mixture = $V + V = 2V \ L$. The number of moles of $NO_3^-$ ions comes from $AgNO_3$. Moles of $AgNO_3$ = $Molarity 	imes Volume = 0.1 \ M 	imes V \ L = 0.1V \ mol$. Since $AgNO_3$ dissociates as $AgNO_3 ightarrow Ag^+ + NO_3^-$,the moles of $NO_3^-$ = $0.1V \ mol$. Concentration of $NO_3^-$ in the mixture = $\frac{	ext{Total moles of } NO_3^-}{	ext{Total volume}} = \frac{0.1V}{2V} = 0.05 \ M$.

When equal volumes of $0.1 \ M \ AgNO_3$ and $0.2 \ M \ NaCl$ are mixed,the concentration of $NO_3^-$ in the resulting mixture will be ............. $M$.

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