Question

К раствору массой 100 г с массовой долей хлороводорода 20% прибавили 100 г раствора карбоната натрия. После этого для полной нейтрализации раствора потребовалось 50 мл с массовой долей гидроксида натрия 25% (плотность раствора 1,03 к/мл). Определите массовую долю карбоната натрия в исходном растворе. 

Answer 1

Calculation of Sodium Carbonate Concentration in the Initial Solution

To determine the mass fraction of sodium carbonate in the initial solution, we need to calculate the amount of hydrochloric acid (HCl) and sodium hydroxide (NaOH) involved in the neutralization reaction.

Given: - Mass of the solution = 100 g - Mass fraction of hydrochloric acid (HCl) = 20% - Mass of the sodium carbonate (Na2CO3) solution added = 100 g - Volume of sodium hydroxide (NaOH) solution required for complete neutralization = 50 mL - Mass fraction of sodium hydroxide (NaOH) = 25% - Density of the sodium hydroxide (NaOH) solution = 1.03 g/mL

Let's calculate the amount of hydrochloric acid (HCl) and sodium hydroxide (NaOH) involved in the reaction.

1. Calculation of the amount of hydrochloric acid (HCl): - Mass of hydrochloric acid (HCl) in the initial solution = Mass of the solution * Mass fraction of HCl - Mass of HCl = 100 g * 20% = 20 g

2. Calculation of the amount of sodium hydroxide (NaOH): - Volume of sodium hydroxide (NaOH) solution in liters = Volume of sodium hydroxide (NaOH) solution / 1000 - Volume of sodium hydroxide (NaOH) solution in liters = 50 mL / 1000 = 0.05 L - Mass of sodium hydroxide (NaOH) = Volume of sodium hydroxide (NaOH) solution * Density of NaOH solution - Mass of NaOH = 0.05 L * 1.03 g/mL = 0.0515 g

3. Calculation of the amount of sodium carbonate (Na2CO3): - Mass of sodium carbonate (Na2CO3) in the added solution = Mass of the sodium carbonate solution * Mass fraction of Na2CO3 - Mass of Na2CO3 = 100 g * X (unknown mass fraction of Na2CO3)

Since the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is 1:1, the moles of HCl reacted will be equal to the moles of NaOH used for neutralization. We can use this information to calculate the unknown mass fraction of sodium carbonate (Na2CO3) in the initial solution.

4. Calculation of the moles of hydrochloric acid (HCl): - Moles of HCl = Mass of HCl / Molar mass of HCl - Molar mass of HCl = 36.46 g/mol (hydrochloric acid) - Moles of HCl = 20 g / 36.46 g/mol = 0.548 mol

5. Calculation of the moles of sodium hydroxide (NaOH): - Moles of NaOH = Moles of HCl (from the neutralization reaction)

6. Calculation of the mass of sodium carbonate (Na2CO3): - Moles of Na2CO3 = Moles of NaOH (from the neutralization reaction) - Mass of Na2CO3 = Moles of Na2CO3 * Molar mass of Na2CO3 - Molar mass of Na2CO3 = 105.99 g/mol (sodium carbonate)

Now, we can calculate the mass fraction of sodium carbonate (Na2CO3) in the initial solution.

7. Calculation of the mass fraction of sodium carbonate (Na2CO3): - Mass fraction of Na2CO3 = Mass of Na2CO3 / Mass of the solution

Let's calculate the values step by step.

Calculation Steps:

1. Mass of hydrochloric acid (HCl): - Mass of HCl = 100 g * 20% = 20 g

2. Mass of sodium hydroxide (NaOH): - Volume of NaOH solution = 50 mL - Volume of NaOH solution in liters = 50 mL / 1000 = 0.05 L - Mass of NaOH = 0.05 L * 1.03 g/mL = 0.0515 g

3. Moles of hydrochloric acid (HCl): - Molar mass of HCl = 36.46 g/mol - Moles of HCl = 20 g / 36.46 g/mol = 0.548 mol

4. Moles of sodium hydroxide (NaOH): - Moles of NaOH = Moles of HCl = 0.548 mol

5. Mass of sodium carbonate (Na2CO3): - Molar mass of Na2CO3 = 105.99 g/mol - Mass of Na2CO3 = 0.548 mol * 105.99 g/mol = 57.98 g

6. Mass fraction of sodium carbonate (Na2CO3): - Mass fraction of Na2CO3 = Mass of Na2CO3 / Mass of the solution - Mass fraction of Na2CO3 = 57.98 g / 200 g = 0.2899

Answer:

The mass fraction of sodium carbonate (Na2CO3) in the initial solution is approximately 28.99%.

Please note that the calculations provided are based on the given information and assumptions.