Finding Normality A solution containing 25.0 mL of oxalic acid required 13.78 mL of 0.041 62 N KMnO4 for titration, according to Reaction E-5. Find the normality and molarity of the oxalic acid.

Question

Finding Normality

A solution containing 25.0 mL of oxalic acid required 13.78 mL of 0.041 62 N [latex]KMnO_{4}[/latex] for titration, according to Reaction E-5. Find the normality and molarity of the oxalic acid.

[latex]5H_{2}C_{2}O_{4} + 2MnO_{4}^{−} + 6H^{+} \rightleftharpoons 2Mn^{2+} + 10CO_{2} + 8H_{2}O[/latex] (E-5)

Accepted Answer

Setting up Equation E-6, we write

[latex]N_{1}V_{1} = N_{2}V_{2}[/latex] (E-6)

[latex]N_{1}[/latex](25.0 mL) = (0.041 62 N)(13.78 mL)

[latex]N_{1}[/latex] = 0.022 94 equiv/L

Because there are two equivalents per mole of oxalic acid in Reaction E-5,

M = [latex]\frac{N}{n} = \frac{0.022 94}{2}[/latex] = 0.011 47 M

Question AE.3: Finding Normality A solution containing 25.0 mL of oxalic ac...

Related Answered Questions

Using Normality How many grams of potassium oxalate should be dissolved in 500.0 mL to make a 0.100 N solution for titration of MnO4^− ? 5H2C2O4 + 2MnO4^− + 6H^+ ⇔ 2Mn^2+ + 10CO2 + 8H2O (E-5) ...

Verified Answer:

Finding Normality Find the normality of a solution containing 6.34 g of ascorbic acid in 250.0 mL if the relevant half-reaction is ...

Verified Answer: