The reaction between vanadium(IV) ions and manganate(VII) ions in acidic solution can be represented by the equation
$$5V^{4+} + MnO_4^{-} + 8H^{+} \rightarrow 5V^{5+} + Mn^{2+} + 4H_2O$$
What volume, in dm³, of 0.020 mol dm⁻³ KMnO₄ is needed to oxidise 0.10 mol of vanadium(IV) ions completely? - AQA - A-Level Chemistry - Question 21 - 2022 - Paper 3
Question 21
The reaction between vanadium(IV) ions and manganate(VII) ions in acidic solution can be represented by the equation
$$5V^{4+} + MnO_4^{-} + 8H^{+} \rightarrow 5V^{... show full transcript
Worked Solution & Example Answer:The reaction between vanadium(IV) ions and manganate(VII) ions in acidic solution can be represented by the equation
$$5V^{4+} + MnO_4^{-} + 8H^{+} \rightarrow 5V^{5+} + Mn^{2+} + 4H_2O$$
What volume, in dm³, of 0.020 mol dm⁻³ KMnO₄ is needed to oxidise 0.10 mol of vanadium(IV) ions completely? - AQA - A-Level Chemistry - Question 21 - 2022 - Paper 3
Step 1
Calculate the moles of KMnO₄ needed to react with vanadium(IV)
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Answer
From the stoichiometry of the reaction, 1 mole of MnO₄⁻ reacts with 5 moles of V^{4+}. Therefore, for 0.10 moles of V^{4+}, the moles of KMnO₄ required are:
n=50.10=0.020 moles
Step 2
Use the concentration of KMnO₄ to find the volume required
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Answer
Using the formula:
extC=Vn
where C is the concentration (0.020 mol/dm³) and n is the number of moles (0.020 moles), we can rearrange the formula to find the volume (V):
