What volume of 6.0 M solution is exactly needed to oxidize 8.0 g of ions to ions, assuming the gets — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

Number of equivalents of $Fe^{2+}$ = $(8.0 / 56) \times 1 = 1/7$ eq. The reduction half-reaction for the oxidizing agent is $NO_3^- + 4H^+ + 3e^- \rightarrow NO + 2H_2O$. Thus, the n-factor of $HNO_3$ is 3. Equivalents of $HNO_3 = M \times n \times V = 6.0 \times 3 \times V(in\ L) = 18V$. By the Law of Equivalence: $18V = 1/7 \Rightarrow V = 1/126\ L \approx 7.936\ mL$.