You are to determine, by titration, the change in oxidation number of a transition metal ion, M^2+, when reacted with acidified potassium manganate(VII) - CIE - A-Level Chemistry - Question 1 - 2014 - Paper 1

To calculate the number of moles of potassium manganate(VII) present in 22.55 cm^3 of FA 1, we use the molarity formula:

ext{Moles} = rac{ ext{Concentration (mol dm}^{-3} ext{) } imes ext{Volume (dm}^{3})}{1000}

From the question, the concentration of FA 1 is 0.0200 mol dm^-3.

Converting the volume to dm^3: $22.55 ext{ cm}^3 = 0.02255 ext{ dm}^3$

Now we can calculate:

$ext{Moles of KMnO}_4 = 0.0200 imes 0.02255 = 0.000451 ext{ mol}$

To calculate the moles of MSO4 in 25.0 cm^3 of FA 2, we apply the same molarity formula:

ext{Moles} = rac{ ext{Concentration (mol dm}^{-3} ext{) } imes ext{Volume (dm}^{3})}{1000}

Given that the concentration of FA 2 is 0.0530 mol dm^-3:

Converting volume: $25.0 ext{ cm}^3 = 0.0250 ext{ dm}^3$

Now calculating: $ext{Moles of MSO}_4 = 0.0530 imes 0.0250 = 0.001325 ext{ mol}$

From the calculation in (i), we found that 0.000451 mol of KMnO4 was present. In the calculations in (ii), we found 0.001325 mol of MSO4 was present. The stoichiometric ratio between KMnO4 and MSO4 can be determined from the balanced equation.

Assuming the reaction is 1:5:

$ext{Moles of MSO}_4 = 5 imes ext{Moles of KMnO}_4 = 5 imes 0.000451 = 0.002255 ext{ mol}$

Based on the calculation in (iii), which suggests that for every 1 mole of KMnO4, 5 moles of MSO4 react. This is consistent with equation 1, which states:

$2KMnO_4 + 10MSO_4 + 8H_2SO_4 → K_2SO_4 + 2MnSO_4 + 5M(SO_4)_2 + 8H_2O$

From this equation, we see that the molar ratio aligns with our calculated moles, confirming our findings.

Using the conclusion from (iv) where the reaction shows that 5 moles of MSO4 produces M in the products,

the oxidation state of M can be calculated from the products formed. In the case of M(SO4)2, the sulfate ion (SO4) carries a -2 charge:

So, for $ext{oxidation state of M} + 2(-2) = 0$

This simplifies to: $ext{oxidation state of M} = +4$

Thus, the oxidation number of the transition metal M in the product is +4.