# Question e5020

Jun 23, 2015

The concentration of the chloride ions will be equal to 0.5 M.

#### Explanation:

The first thing that you need to notice is that both compounds contain chloride ions, and that both are soluble in aqueous solution.

This means that they will dissociate to form

$N a C {l}_{\left(a q\right)} \to N {a}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-}$

and

$B a C {l}_{2 \left(a q\right)} \to B {a}_{\left(a q\right)}^{2 +} + \textcolor{red}{2} C {l}_{\left(a q\right)}^{-}$

So, every mole of sodium chloride produces 1 mole of chloride ions, and every mole of barium chloride produces $\textcolor{red}{2}$ moles of chloride ions in solution.

Use the molarities and volumes of the two solutions to determine how many moles of each compound you have present

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{N a C l} = \text{0.3 M" * 300 * 10^(-3)"L" = "0.09 moles}$ $N a C l$

and

${n}_{B a C {l}_{2}} = \text{0.4 M" * 200 * 10^(-3)"M" = "0.08 moles}$ $B a C {l}_{2}$

According to the dissociation equations, you have

0.09cancel("moles"NaCl) * ("1 mole "Cl^(-))/(1cancel("mole"NaCl)) = "0.09 moles" $C {l}^{-}$

and

0.08cancel("moles"BaCl_2) * (color(red)(2)"moles "Cl^(-))/(1cancel("mole"BaCl_2)) = "0.16 moles" $C {l}^{-}$

The total number of moles of chloride ions will thus be

${n}_{\text{total" = 0.09 + 0.16 = "0.25 moles }}$$C {l}^{-}$

The total volume of the solution will be

${V}_{\text{total}} = {V}_{N a C l} + {V}_{B a C {l}_{2}}$

${V}_{\text{total" = 300 + 200 = "500 mL}}$

Therefore, the molarity of the chloride ions will be

[Cl^(-)] = n_"total"/V_"total" = "0.25 moles"/(500 * 10^(-3)"L") = color(green)("0.5 M")#