Calculate the concentration of NaOH if 25 ml of it is titrated against 23.7 ml of a 6.3g ethanedioic solution

To determine the concentration of sodium hydroxide (NaOH) when titrated against a solution of ethanedioic acid (also known as oxalic acid), we need to follow a systematic approach. This involves understanding the reaction between NaOH and ethanedioic acid, calculating the moles of ethanedioic acid present, and then using that information to find the concentration of NaOH.

Step 1: Understanding the Reaction

The balanced chemical equation for the reaction between sodium hydroxide and ethanedioic acid is:

H₂C₂O₄ + 2 NaOH → Na₂C₂O₄ + 2 H₂O

This equation tells us that one mole of ethanedioic acid reacts with two moles of sodium hydroxide. Therefore, the stoichiometry of the reaction is crucial for our calculations.

Step 2: Calculating Moles of Ethanedioic Acid

First, we need to find the number of moles of ethanedioic acid in the solution. The mass of ethanedioic acid is given as 6.3 grams. The molar mass of ethanedioic acid (H₂C₂O₄) is approximately 90.03 g/mol. We can calculate the moles using the formula:

Moles = Mass (g) / Molar Mass (g/mol)

Substituting the values:

Moles of ethanedioic acid = 6.3 g / 90.03 g/mol ≈ 0.0700 moles

Step 3: Relating Moles of Ethanedioic Acid to NaOH

From the balanced equation, we know that 1 mole of ethanedioic acid reacts with 2 moles of NaOH. Therefore, the moles of NaOH required for the reaction can be calculated as follows:

Moles of NaOH = 2 × Moles of ethanedioic acid

Moles of NaOH = 2 × 0.0700 moles = 0.1400 moles

Step 4: Finding the Concentration of NaOH

Now that we know the moles of NaOH, we can find its concentration. The concentration (C) is defined as the number of moles of solute divided by the volume of solution in liters:

Concentration (C) = Moles of NaOH / Volume of NaOH solution (L)

The volume of NaOH solution is given as 25 mL, which we need to convert to liters:

Volume of NaOH = 25 mL × (1 L / 1000 mL) = 0.025 L

Now we can calculate the concentration:

C = 0.1400 moles / 0.025 L = 5.6 M

Final Result

The concentration of the sodium hydroxide solution is approximately 5.6 M. This means that in every liter of this NaOH solution, there are 5.6 moles of sodium hydroxide present.