20 ml of a solution containing Na2CO3 and NaOH was first tutrated using phenofthlien against N/10HCl. The end point reached when 17. 5 ml of the acid was used up after this methyl orange was added and 2.5 ml of HCl was again required to reach the next end point. find out the amount of NaOH and Na2CO3 in the solution

To determine the amounts of sodium carbonate (Na2CO3) and sodium hydroxide (NaOH) in the solution, we can break the problem down into a few logical steps. We will use the information from the titration with hydrochloric acid (HCl) and the indicators used to identify the endpoints. Let's go through the calculations step by step.

Understanding the Reaction

When Na2CO3 and NaOH are titrated with HCl, they react in different ways:

• Na2CO3 reacts with HCl to form NaCl, water, and carbon dioxide:
• Na2CO3 + 2 HCl → 2 NaCl + H2O + CO2↑
• NaOH reacts with HCl to form NaCl and water:
• NaOH + HCl → NaCl + H2O

Analyzing the Titration Data

In the first titration, 17.5 ml of N/10 HCl was used to reach the endpoint with phenolphthalein. This endpoint indicates that all the NaOH and half of the Na2CO3 have reacted. In the second titration, after adding methyl orange, an additional 2.5 ml of HCl was required to reach the next endpoint. This endpoint indicates that the remaining Na2CO3 has reacted completely.

Calculating Moles of HCl Used

First, we need to calculate the moles of HCl used in both titrations:

• Volume of HCl in the first titration: 17.5 ml = 0.0175 L
• Volume of HCl in the second titration: 2.5 ml = 0.0025 L
• Total volume of HCl used: 0.0175 L + 0.0025 L = 0.0200 L

The concentration of HCl is N/10, which is 0.1 N. Therefore, the total moles of HCl used is:

Moles of HCl = Concentration × Volume

Moles of HCl = 0.1 mol/L × 0.0200 L = 0.0020 moles

Distributing Moles to Na2CO3 and NaOH

From the titration data, we can deduce how the moles of HCl correspond to Na2CO3 and NaOH:

• In the first titration (17.5 ml), all NaOH and half of Na2CO3 reacted:
• Let x be the moles of NaOH and y be the moles of Na2CO3.
• The reaction gives us the equation: x + y/2 = 0.0020 (1)
• In the second titration (2.5 ml), the remaining Na2CO3 reacted:
• This means that y/2 moles of HCl were used, leading to the equation: y/2 = 0.0025 × 0.1 = 0.00025 (2)

Solving the Equations

From equation (2), we can find y:

y = 0.00025 × 2 = 0.0005 moles

Now substituting y back into equation (1):

x + 0.0005/2 = 0.0020

x + 0.00025 = 0.0020

x = 0.0020 - 0.00025 = 0.00175 moles

Final Amounts of NaOH and Na2CO3

Now that we have the moles of NaOH and Na2CO3, we can convert these to grams if needed:

• Molar mass of NaOH = 40 g/mol
• Molar mass of Na2CO3 = 106 g/mol

Calculating the mass:

• Mass of NaOH = 0.00175 moles × 40 g/mol = 0.070 g
• Mass of Na2CO3 = 0.0005 moles × 106 g/mol = 0.053 g

In summary, the solution contains approximately 0.070 g of NaOH and 0.053 g of Na2CO3. This methodical approach allows us to understand the relationships between the reactants and the titration process effectively.