A solution contains Na2CO3 and NaHCO3 . 10 ml
of this solution required 2.5ml of 0.1 M H2SO4 for
neutralizing using phenolphthalein as indicator.
Further 2.5 ml of 0.2 M H2SO4 is needed when
methyl orange is used as indicator after the first end
point. what is the mass of Na2CO3 & NaHCO3 in
one lit of solution -

To determine the mass of Na2CO3 (sodium carbonate) and NaHCO3 (sodium bicarbonate) in one liter of the solution, we need to analyze the neutralization reactions that occur with sulfuric acid (H2SO4) using the two indicators, phenolphthalein and methyl orange. Each indicator signifies a different endpoint in the titration process, indicating the presence of different species in the solution.

Understanding the Reactions

First, let's break down the neutralization reactions:

• Na2CO3 reacts with H2SO4 to produce Na2SO4, CO2, and H2O:
• Na2CO3 + H2SO4 → Na2SO4 + CO2 + H2O
• NaHCO3 reacts with H2SO4 to produce Na2SO4, CO2, and H2O:
• NaHCO3 + H2SO4 → Na2SO4 + CO2 + H2O

In the first part of the titration, phenolphthalein is used, which indicates the endpoint when all the Na2CO3 has reacted. The second part, using methyl orange, indicates the endpoint when NaHCO3 is fully neutralized.

Calculating Moles of H2SO4 Used

Now, let’s calculate the moles of H2SO4 used in both steps:

• For the first endpoint with phenolphthalein:
• Volume of H2SO4 = 2.5 mL = 0.0025 L
• Concentration of H2SO4 = 0.1 M
• Moles of H2SO4 = Volume × Concentration = 0.0025 L × 0.1 mol/L = 0.00025 moles
• For the second endpoint with methyl orange:
• Volume of H2SO4 = 2.5 mL = 0.0025 L
• Concentration of H2SO4 = 0.2 M
• Moles of H2SO4 = Volume × Concentration = 0.0025 L × 0.2 mol/L = 0.0005 moles

Total Moles of H2SO4

Now, we can find the total moles of H2SO4 used:

• Total moles of H2SO4 = 0.00025 moles + 0.0005 moles = 0.00075 moles

Relating Moles of H2SO4 to Moles of Na2CO3 and NaHCO3

From the reactions, we know:

• 1 mole of Na2CO3 reacts with 1 mole of H2SO4.
• 1 mole of NaHCO3 reacts with 1 mole of H2SO4.

Let’s denote:

• x = moles of Na2CO3
• y = moles of NaHCO3

From the reactions, we can set up the following equations:

• x + y = 0.00075 (total moles of H2SO4)

Next, we need to consider the stoichiometry of the reactions. Since Na2CO3 is fully neutralized first, we can assume that all of it reacts with the first 0.00025 moles of H2SO4:

• x = 0.00025

Substituting this back into the first equation:

• 0.00025 + y = 0.00075
• y = 0.00075 - 0.00025 = 0.0005

Calculating Masses

Now that we have the moles of each compound, we can calculate the mass:

• Molar mass of Na2CO3 = 2(23) + 12 + 3(16) = 106 g/mol
• Molar mass of NaHCO3 = 23 + 1 + 12 + 3(16) = 84 g/mol

Now, we can find the mass in 1 L of solution:

• Mass of Na2CO3 = moles × molar mass = 0.00025 moles × 106 g/mol = 0.0265 g
• Mass of NaHCO3 = moles × molar mass = 0.0005 moles × 84 g/mol = 0.042 g

Final Results

To summarize, in one liter of the solution, the masses of Na2CO3 and NaHCO3 are:

• Na2CO3: 0.0265 g
• NaHCO3: 0.042 g

This analysis shows how the stoichiometry of acid-base reactions can help us determine the composition of a solution based on titration data. If you have any further questions or need clarification on any part of this process, feel free to ask!