Mn²⁺ is oxidised by peroxydisulphate to permanganate as Mn²⁺ + S₂O₈²⁻ → MnO₄⁻ + SO₄²⁻. 150 mL of 0.35 M S₂O₈²⁻ require 21 mL of a solution of Mn²⁺. What is the molarity of Mn²⁺ sol... Mn²⁺ is oxidised by peroxydisulphate to permanganate as Mn²⁺ + S₂O₈²⁻ → MnO₄⁻ + SO₄²⁻. 150 mL of 0.35 M S₂O₈²⁻ require 21 mL of a solution of Mn²⁺. What is the molarity of Mn²⁺ solution? Read more

Steps to Solve

1. Determine Moles of S₂O₈²⁻ First, calculate the moles of peroxydisulfate ($S_2O_8^{2-}$) used in the reaction. Use the formula: $$ \text{Moles} = \text{Molarity} \times \text{Volume (in L)} $$ For $150 , \text{mL}$ of $0.35 , \text{M}$, convert volume from mL to L: $$ 150 , \text{mL} = 0.150 , \text{L} $$ Now calculate: $$ \text{Moles of } S_2O_8^{2-} = 0.35 , \text{mol/L} \times 0.150 , \text{L} $$ $$ = 0.0525 , \text{mol} $$

2. Stoichiometry of the Reaction From the balanced equation: $$ \text{Mn}^{2+} + S_2O_8^{2-} \rightarrow \text{MnO}_4^{-} + \text{SO}_4^{2-} $$ The stoichiometric ratio is 1:1, meaning 1 mole of $S_2O_8^{2-}$ reacts with 1 mole of $Mn^{2+}$.

3. Calculate Moles of Mn²⁺ Since the ratio is 1:1, the moles of $Mn^{2+}$ will be equal to the moles of $S_2O_8^{2-}$: $$ \text{Moles of } Mn^{2+} = 0.0525 , \text{mol} $$

4. Calculate Molarity of Mn²⁺ Use the volume of the $Mn^{2+}$ solution, which is $21 , \text{mL}$ (convert to L): $$ 21 , \text{mL} = 0.021 , \text{L} $$ Now calculate the molarity: $$ \text{Molarity} = \frac{\text{Moles}}{\text{Volume (in L)}} $$ $$ \text{Molarity of } Mn^{2+} = \frac{0.0525 , \text{mol}}{0.021 , \text{L}} $$

5. Final Calculation Perform the calculation: $$ \text{Molarity of } Mn^{2+} = 2.5 , \text{M} $$