Question

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

Answer 1

Answer:

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:

$M_2=\frac{n_2}{V_2}\\\\n_2=M_2\times V_2=0.145\times 0.018 = 2.61\times 10^{-3}\ mol$

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

So, the complete balanced equation is given below.

$6Fe^{2+}(aq)+Cr_2O_7^{2-}(aq)+14H^+(aq)\to 6Fe^{3+}(aq)+2Cr^{3+}(aq)+7H_2O$

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)

                                          = $6\times 2.61\times 10^{-3}=0.016\ mol\ Fe^{2+}$

So, 0.016 moles of iron(II) is needed. Therefore, $n_1=0.016\ mol$

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

Molarity = Moles ÷ Volume

$M_1=\frac{n_1}{V_1}\\\\M_1=\frac{0.016\ mol}{0.025\ L}=0.64\ M$

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