A 2.51 times 10^-2 L sample of a solution of Cu^ requires 3.30 times 10^-2 L of 0.132 M KMnO_4 solution to reach the equivalence point. The products of the reaction are Cu^2 and Mn^2 . What is the starting concentration of the Cu^ solution

0.868M Cu⁺

Explanation:

The reaction is:

KMnO₄ + 5Cu⁺ + 8H⁺ → 5Cu²⁺ + Mn²⁺ + K⁺ + 4H₂O

Where 1 mol of KMnO₄ reacts with 5 moles of Cu⁺

3.30x10⁻²L of 0.132M KMnO₄ are:

3.30x10⁻²L ×  (0.132mol KMnO₄ / L) = 4.356x10⁻³mol KMnO₄

As 1 mol reacts with 5 moles of Cu⁺, moles of Cu⁺ are:

4.356x10⁻³mol KMnO₄ × (5mol Cu⁺ / 1mol KMnO₄) = 0.02178moles Cu⁺

As these moles are in 2.51x10⁻²L, the starting concentration of the Cu⁺ solution is:

0.02178moles Cu⁺ / 2.51x10⁻²L = 0.868M Cu⁺

- lead + phosphoric acid = hydrogen + lead(ii) phosphate