Question

What is the molar concentration of Fe2+ ion in an aqueous solution if 27.50 mL of 0.118 M KBrO3 is required for complete reaction with 13.00 mL of the Fe2+ solution? The net ionic equation is:
6Fe2+(aq)+BrO3−(aq)+6H+(aq)→6Fe3+(aq)+Br−(aq)+3H2O(l).

Answer 1

First, we have to calculate the amount in moles of $BrO_3^-$ reacted:

$n(BrO_3^-)=C(BrO_3^-)*V(BrO_3^-)=0.118M*0.02750L=3.245*10^{-3}mol$

According to the given equation, the amount of the iron(II) cations required is six time higher then that of the $BrO_3^-$ anions:

$n(Fe^{2+})=6*n(BrO_3^-)=6*3.245*10^{-3}mol=0.01947mol$

Now, it the right time to calculate the molar concentration of the iron(II) cations in the starting mixture:

$C(Fe^{2+})=\frac{n(Fe^{2+})}{V(Fe^{2+})}=\frac{0.01947mol}{0.01300L}=1.498M$

Thus, it was 1.498 M water solution of some iron(II) salt.