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The concentration of HCl can be determined by titration with a primary standard such as Na2CO3 (sodium carbonate). The titration reaction is CO3(2-) + 2H+ ---> H2O + CO2 What is the concentration of HCl if 35.7 mL of HCl was needed to to react with 0.521 g of Na2CO3 to reach the end point?

Respuesta :

Answer:

The concentration of HCl is 0.275 M

Explanation:

Step 1: Data given

Mass of Na2CO3 = 0.521 grams

Molar mass of Na2CO3 = 105.99 g/mol

Volume of HCl = 35.7 mL

Step 2: The balanced equation:

2HCl + Na2CO3 → 2NaCl + CO2 + H2O

CO3(2-) + 2H+ ---> H2O + CO2

Step 3: Calculate moles of Na2CO3

Moles Na2CO3 = mass Na2CO3 / Molar mass Na2CO3

Moles Na2CO3 = 0.521 grams / 105.99 g/mol

Moles Na2CO3 = 0.0049 moles

Step 4: Calculate moles of HCl

For 1 mol of Na2CO3 consumed, we need 2 moles of HCl

moles of HCL = 2 * 0.0049 = 0.0098 moles HCl

Step 5: Calculate molarity of HCl

Molarity = number of moles of HCl / Volume (liters)

Molarity = 0.0098 moles / 0.0357 L

Molarity = 0.275 M

The concentration of HCl is 0.275 M

Oseni

The concentration of HCl if 35.7 mL of HCl was needed to react with 0.521 g of Na2CO3 to reach the endpoint would be 0.27 mol/L or M

Stoichiometric calculations

From the equation of the reaction, the mole ratio of HCl to Na2CO3 is 2:1

Mole of 0.521 g Na2CO3 = 0.521/105.99

                                             = 0.0049 moles

Equivalent mole of HCl = 0.0049 x 2 = 0.0098 moles

Concentration of 0.0098 moles, 35.7 mL HCl = 0.0098/0.0357

                                               = 0.27 mol/L or M

More on stoichiometric calculations can be found here: https://brainly.com/question/8062886