a mixture of na2c2o4 and h2c2o4 requires 100ml of 0.1M KMno4 for complete neutralization .The same mixture on reaction with a base requires 50ml of 0.2 M NaoH solution . what is the moles of oxilate ions in the mxture?

To determine the moles of oxalate ions in the mixture of sodium oxalate (Na2C2O4) and oxalic acid (H2C2O4), we need to analyze the reactions involved with potassium permanganate (KMnO4) and sodium hydroxide (NaOH). Both reactions will help us find the total moles of oxalate ions present in the mixture.

Understanding the Reactions

First, let's break down the reactions that occur with KMnO4 and NaOH. Potassium permanganate acts as a strong oxidizing agent in acidic conditions, while sodium hydroxide provides a basic environment for neutralization.

Reaction with KMnO4

The balanced reaction of KMnO4 with oxalate ions (C2O4^2-) in an acidic medium can be represented as follows:

• 2 KMnO4 + 5 C2O4^2- + 6 H+ → 2 Mn^2+ + 10 CO2 + 8 H2O

From this equation, we see that 2 moles of KMnO4 react with 5 moles of oxalate ions. Since we used 100 mL of 0.1 M KMnO4, we can calculate the moles of KMnO4 used:

• Moles of KMnO4 = Concentration × Volume = 0.1 mol/L × 0.1 L = 0.01 moles

Now, using the stoichiometry from the balanced equation, we can find the moles of oxalate ions:

• From the reaction, 2 moles of KMnO4 react with 5 moles of C2O4^2-.
• Thus, 0.01 moles of KMnO4 will react with (5/2) × 0.01 = 0.025 moles of C2O4^2-.

Reaction with NaOH

Next, let's consider the reaction with sodium hydroxide. The neutralization of oxalic acid (H2C2O4) with NaOH can be represented as:

• H2C2O4 + 2 NaOH → Na2C2O4 + 2 H2O

In this case, each mole of oxalic acid reacts with 2 moles of NaOH. We used 50 mL of 0.2 M NaOH, so we can calculate the moles of NaOH used:

• Moles of NaOH = Concentration × Volume = 0.2 mol/L × 0.05 L = 0.01 moles

Since 2 moles of NaOH are required for each mole of oxalic acid, the moles of oxalic acid that reacted can be calculated as:

• Moles of H2C2O4 = Moles of NaOH / 2 = 0.01 moles / 2 = 0.005 moles.

Calculating Total Moles of Oxalate Ions

Now, we can find the total moles of oxalate ions in the mixture. The total contribution of oxalate ions comes from both sodium oxalate and oxalic acid:

• From sodium oxalate (Na2C2O4), we have 0.025 moles of C2O4^2-.
• From oxalic acid (H2C2O4), we have 0.005 moles of C2O4^2- (since each mole of H2C2O4 provides one mole of C2O4^2-).

Adding these together gives:

• Total moles of oxalate ions = 0.025 moles + 0.005 moles = 0.030 moles.

Final Result

Therefore, the total moles of oxalate ions in the mixture is 0.030 moles.