The titration of 25.0 mL of an iron(II) solution required 18.0 mL of a 0.145 M solution of dichromate to reach the equivalence point. What is the molarity of the iron(II) solution

0.64 M

Explanation:

Given:

Volume of iron(II) solution (V₁) = 25.0 mL = 0.025 L

Molarity of iron(II) solution (M₁) = ?

Number of moles of iron(II) solution (n₁) = ?

Volume of dichromate solution (V₂) = 18.0 mL = 0.018 L

Molarity of dichromate solution (M₂) = 0.145 M

Number of moles of dichromate solution (n₂) = ?

Molarity is equal to the ratio of moles and volume.

So, molarity of dichromate solution is given as:

Now, let us write the complete balanced reaction for the given situation.

So, the complete balanced equation is given below.

From the equation, it is clear that, 1 mole of dichromate is required for 6 moles of iron(II) solution.

So, using unitary method, we find the number of moles of iron(II) solution.

1 mole of dichromate = 6 moles of iron(II)

∴ n₂ moles of dichromate = 6n₂ moles of iron(II)

                                          =

So, 0.016 moles of iron(II) is needed. Therefore,

Now, molarity of iron(II) solution is given as:

Molarity = Moles ÷ Volume

Therefore, the molarity of the iron(II) solution is 0.64 M.

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