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