Iron fluoride (FeF2) dissociates according to the following equation:
FeF2(s) ⇌ Fe^2+(aq) + 2 F^–(aq)
A sample of iron fluoride (FeF2) is dissolved in water, and a saturated solution is obtained. The [Fe^2+] is measured to be 2.05 × 10^–4 mol/L.
What is the concentration of fluoride (F^–) ions?
It has to be twice that.
It's not 2.05 x 10^-4 mol/L
Bobpursley has to be right it must be double!
the peeps above me r right! just do wha dey say:
(2.05 × 10^–4) x 2 = 4.10 × 10^–4 mol/L
hope dis helps <3
To find the concentration of fluoride ions (F^-), we need to use the balanced equation for the dissociation of iron fluoride (FeF2):
FeF2(s) ⇌ Fe^2+(aq) + 2 F^–(aq)
From the balanced equation, we can see that for every 1 mol of FeF2 that dissociates, 2 mol of fluoride ions (F^-) are produced. This means that the concentration of F^- ions will be twice the concentration of Fe^2+ ions.
Given that the concentration of Fe^2+ ions ([Fe^2+]) is 2.05 × 10^–4 mol/L, we can calculate the concentration of F^- ions ([F^-]) as follows:
[F^-] = 2 × [Fe^2+]
[F^-] = 2 × (2.05 × 10^–4 mol/L)
[F^-] = 4.10 × 10^–4 mol/L
Therefore, the concentration of fluoride ions (F^-) is 4.10 × 10^–4 mol/L.