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mezya [45]
2 years ago
13

You mix 500.0 mL of 0.250 M iron(III) chloride solution with 425.0 mL of 0.350 M barium chloride solution. Assuming the volumes

are additive, what is the molarity of chloride ion in the mixture.
Chemistry
1 answer:
12345 [234]2 years ago
5 0

Answer:

M=0.727M

Explanation:

Hello,

In this case, since iron (III) chloride (FeCl3) and barium chloride (BaCl2) are both chloride-containing compounds, we can compute the moles of chloride from each salt, considering the concentration and volume of the given solutions, and using the mole ratio that is 1:3 and 1:2 for the compound to chlorine:

n_{Cl^-}=0.50L*0.250\frac{molFeCl_3}{L}*\frac{3molCl^-}{1molFeCl_3}=0.375molCl^-  \\\\n_{Cl^-}=0.425L*0.350\frac{molBaCl_2}{L}*\frac{2molCl^-}{1molBaCl_2}=0.2975molCl^-

So the total mole of chloride ions:

N_{Cl^-}=0.2975mol+0.375mol=0.6725molCl^-

And the total volume by adding the volume of each solution in L:

V=0.500L+0.425L=0.925L

Finally, the molarity turns out:

M=\frac{0.6725molCl^-}{0.925L}\\ \\M=0.727M

Best regards.

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