A solution of (NH4)2SO4·FeSO4·6H2O (hydrated ammonium iron(II) sulfate), used as a primary standard, was prepared by dissolving 8.82 g of the crystals in dilute sulfuric acid and making the solution up to exactly 250 cm³ with deionized water - Leaving Cert Chemistry - Question 1 - 2017

1. The primary standard must be of high purity and stable, which means it should not decompose or react with air and moisture.
2. It should have a known and reliable molar mass to ensure accurate calculations during titrations.

Dilute sulfuric acid is added to ensure that the iron(II) ions are present in a suitable form for titration. It prevents oxidation of Fe²⁺ to Fe³⁺ by atmospheric oxygen. Additionally, the acidic environment stabilizes Fe²⁺ ions, avoiding precipitation and maintaining the solution's integrity.

Adding dilute sulfuric acid to each 25.0 cm³ portion during titrations ensures that the reaction proceeds efficiently by maintaining an acidic environment, which prevents the oxidation of Fe²⁺ during the titration and is critical for accurate measurement of endpoints.

The procedure begins by rinsing the burette with potassium manganate(VII) solution and filling it without air bubbles. The appropriate amount of standard solution is placed in a conical flask with some dilute sulfuric acid. The manganate solution is added slowly while constantly swirling the flask, ensuring the sample is mixed well. The endpoint is detected by the first permanent pink color that persists after swirling.

The correct endpoint is indicated by a persistent pale pink color in the solution, which shows that all Fe²⁺ ions have been oxidized, and the addition of the manganate(VII) solution has caused a slight excess of MnO4⁻.

To find the number of moles of Fe²⁺, use the formula:

$n = \frac{m}{M}$

Where:

• m = 0.882 g (mass of iron(II) sulfate)
• M = 392 g/mol (molar mass of hydrated iron(II) sulfate)

Substituting: $n = \frac{0.882 \, g}{392 \, g/mol} = 0.00225 \, moles$

Using the reaction stoichiometry, 1 mole of MnO4⁻ reacts with 5 moles of Fe²⁺. Therefore, the moles of potassium manganate(VII) can be calculated from:

$\text{Moles of MnO}_4^- = \frac{1}{5} \times \text{Moles of Fe}^{2+}$

Using the previously calculated moles of Fe²⁺: $\text{Moles of MnO}_4^- = \frac{0.00225}{5} = 0.00045 \, moles$

Molarity (M) is calculated using:

$M = \frac{n}{V}$

Where:

• n = 0.00045 moles (moles of KMnO4)
• V = 0.0205 L (volume of solution in L)

Thus, $M = \frac{0.00045 \, moles}{0.0205 \, L} = 0.022 M$