The titration of 25.0 mL of an iron(II) solution required 18.0 mL of a 0.265 M solution of dichromate to reach the equivalence point. What is the molarity of the iron(II) solution?
(18ml)(.265M)/1 mole=(25ml)(xM)/6mole
xM=1.14M
6Fe^2+ + Cr2O7^2- ==> 2Cr^3+ + 6Fe^3+
mols Cr2O7^2- = M x L = ?
mols Fe^2+ = 6 x that.
M Fe^2+ = mols Fe^2+/L Fe^2+
(Note: I didn't balanced the equation completely; actually we need only the redox part balanced.
Well, that's quite the chemical puzzle you've got there. To find the molarity of the iron(II) solution, we can start by setting up the balanced chemical equation:
6Fe^2+(aq) + Cr2O7^-(aq) + 14H+(aq) --> 6Fe^3+(aq) + 2Cr^3+(aq) + 7H2O(l)
From this equation, we can see that it takes 6 moles of Fe^2+ ions reacting with 1 mole of dichromate ions (Cr2O7^-) to reach the equivalence point.
Given that the volume of the dichromate solution used is 18.0 mL and the molarity is 0.265 M, we can calculate the moles of Cr2O7^- used:
moles of Cr2O7^- = volume (L) x molarity
= 0.0180 L x 0.265 mol/L
= 0.00477 mol
Since the stoichiometric ratio of Fe^2+ to Cr2O7^- is 6:1, we can determine the moles of Fe^2+ present in the iron(II) solution:
moles of Fe^2+ = (moles of Cr2O7^-) / 6
= 0.00477 mol / 6
= 0.000795 mol
Now, let's calculate the molarity of the iron(II) solution:
molarity of Fe^2+ solution = moles of Fe^2+ / volume (L)
= 0.000795 mol / 0.0250 L
= 0.0318 M
So, the molarity of the iron(II) solution is approximately 0.0318 M. But enough with the stoichiometry, chemistry can be such a balancing act sometimes!
To calculate the molarity of the iron(II) solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between iron(II) ions (Fe2+) and dichromate ions (Cr2O7^2-).
The balanced chemical equation for the reaction is:
6 Fe2+ + Cr2O7^2- + 14 H+ -> 6 Fe3+ + 2 Cr3+ + 7 H2O
From the equation, we can see that the ratio of moles of Fe2+ to moles of Cr2O7^2- is 6:1.
Given that 18.0 mL of a 0.265 M solution of dichromate (Cr2O7^2-) was required to reach the equivalence point, we can use this information to calculate the number of moles of Cr2O7^2-:
Moles of Cr2O7^2- = volume (in L) x molarity
= 0.018 L x 0.265 mol/L
= 0.00477 mol
Since the ratio of moles of Fe2+ to moles of Cr2O7^2- is 6:1, the number of moles of Fe2+ can be calculated as:
Moles of Fe2+ = 6 x Moles of Cr2O7^2-
= 6 x 0.00477 mol
= 0.0286 mol
Now, we can calculate the molarity of the iron(II) solution by dividing the number of moles of Fe2+ by the volume of the iron(II) solution in liters:
Molarity of Fe2+ solution = Moles of Fe2+ / Volume of Fe2+ solution (in L)
= 0.0286 mol / 0.0250 L
= 1.14 M
Therefore, the molarity of the iron(II) solution is 1.14 M.