What is the molar concentration of chloride i | vedclass

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(A) The dissociation reactions are:$KCl \rightarrow K^{+} + Cl^{-}$$BaCl_2 \rightarrow Ba^{2+} + 2Cl^{-}$For $KCl$ solution:$Moles \ of \ Cl^{-} = M \

Vedclass · Accepted Answer

$5.0$

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(A) The dissociation reactions are:$KCl ightarrow K^{+} + Cl^{-}$$BaCl_2 ightarrow Ba^{2+} + 2Cl^{-}$For $KCl$ solution:$Moles \ of \ Cl^{-} = M 	imes V = 3.0 \ M 	imes 0.3 \ L = 0.9 \ mol$For $BaCl_2$ solution:$Moles \ of \ Cl^{-} = 2 	imes M 	imes V = 2 	imes 4.0 \ M 	imes 0.2 \ L = 1.6 \ mol$Total moles of $Cl^{-} = 0.9 \ mol + 1.6 \ mol = 2.5 \ mol$Total volume of solution = $300 \ mL + 200 \ mL = 500 \ mL = 0.5 \ L$$[Cl^{-}] = \frac{	ext{Total moles of } Cl^{-}}{	ext{Total volume in } L} = \frac{2.5 \ mol}{0.5 \ L} = 5.0 \ M$

What is the molar concentration of chloride ion in a solution made by adding $300 \ mL$ of $3.0 \ M$ $KCl$ solution to $200 \ mL$ of $4.0 \ M$ $BaCl_2$ solution? $........ \ M$

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