Question

in the laboratory you dissolve 18.6 g of iron(ii) bromide in a volumetric flask and add water to a total volume of 500 ml. what is the molarity of the solution? molarity = what is the concentration of the iron(ii) cation? concentration = m what is the concentration of the bromide anion? concentration = m use the references to access important values if needed for this question.

Explanation:

Step1: Calculate moles of iron(II) bromide

The molar - mass of iron(II) bromide ($FeBr_2$) is $M_{FeBr_2}=55.85 + 2\times79.90=215.65\ g/mol$. The number of moles of $FeBr_2$, $n=\frac{m}{M}$, where $m = 18.6\ g$. So, $n=\frac{18.6\ g}{215.65\ g/mol}\approx0.0863\ mol$.

Step2: Calculate molarity of the solution

The volume of the solution $V = 500\ mL=0.5\ L$. The molarity formula is $M=\frac{n}{V}$. So, $M=\frac{0.0863\ mol}{0.5\ L}=0.173\ M$.

Step3: Determine the concentration of $Fe^{2 + }$ ions

In $FeBr_2$, the dissociation equation is $FeBr_2(s)
ightarrow Fe^{2 + }(aq)+2Br^{-}(aq)$. For every mole of $FeBr_2$ that dissociates, 1 mole of $Fe^{2 + }$ is produced. So the concentration of $Fe^{2 + }$ is equal to the molarity of $FeBr_2$, which is $0.173\ M$.

Step4: Determine the concentration of $Br^{-}$ ions

For every mole of $FeBr_2$ that dissociates, 2 moles of $Br^{-}$ are produced. So the concentration of $Br^{-}$ is $2\times$ the molarity of $FeBr_2$. $[Br^{-}]=2\times0.173\ M = 0.346\ M$.

Answer:

Molarity of the solution: $0.173\ M$
Concentration of iron(II) cation: $0.173\ M$
Concentration of bromide anion: $0.346\ M$