In the laboratory a student combines 44.0 mL of a 0.403 M cobalt(II) fluoride solution with...

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In the laboratory a student combines 44.0 mL of a 0.403 M cobalt(II) fluoride solution with...

In the laboratory a student combines 44.0 mL of a 0.403 M cobalt(II) fluoride solution with 17.5 mL of a 0.530 M iron(III) fluoride solution.

What is the final concentration of fluoride anion ?

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Answer 1

Answer #1

Volume of CoF₂ = 44 mL = 0.044 L

Molarity of CoF₂ = 0.403 M

Moles of CoF₂ = Molarity * Volume = 0.403 M *0.044 L = 0.017732 moles

mol of F^- in CoF₂ = 2* moles of CoF₂ = 2 * 0.017732 = 0.035464 moles

Similarly moles of FeF₃ = 0.0175 L * 0.53 M = 0.009275 moles

Moles of F^- in FeF₃ = 3* mol of FeF₃ = 3* 0.009275 = 0.027825 mol

Total moles of F^- = 0.035464 moles + 0.027825 mol = 0.063289 mol

Total volume of solution = 0.044 L + 0.0175 L = 0.0615 L

[F^-] = concentration in fluoride anion = Moles/ volume = 0.063289 mol / 0.0615 L = 1.03 M

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Answer 2

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In the laboratory a student combines 44.0 mL of a 0.403 M cobalt(II) fluoride solution with...

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