If 0.752 g of pure sodium carbonate was dissolved in water and the solution titrated with 25.90 mL of hydrochloric acid to a methyl orange end point, calculate the molarity of the hydrochloric acid solution. (Hint: This process takes the carbonate ion to carbon dioxide.)

Do I find the moles of sodium carbonate and change it to hydrochloric acid and then put that over the L of hydrochloric acid?

Thanks:)

3 answers

exactly. Good work.

THANKS:)

Yes, you're on the right track! To find the molarity of the hydrochloric acid solution, you need to convert the moles of sodium carbonate to moles of hydrochloric acid and then divide it by the volume of hydrochloric acid.

Here are the steps to calculate the molarity of the hydrochloric acid solution:

1. Calculate the number of moles of sodium carbonate (Na2CO3) using its molar mass. The molar mass of Na2CO3 is Na (22.99 g/mol) + C (12.01 g/mol) + 3O (16.00 g/mol) = 105.99 g/mol. Therefore, 0.752 g of Na2CO3 is equivalent to:

(0.752 g / 105.99 g/mol) = 0.00709 mol of Na2CO3

2. Since the reaction takes the carbonate ion (CO3²-) to form carbon dioxide (CO2) and water (H2O), the balanced chemical equation is:

Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O

3. From the balanced equation, we can see that each mole of Na2CO3 reacts with 2 moles of HCl. Therefore, the number of moles of HCl is:

0.00709 mol Na2CO3 * (2 mol HCl / 1 mol Na2CO3) = 0.01418 mol HCl

4. Finally, calculate the molarity (M) of the hydrochloric acid solution using the formula:

Molarity = Moles of solute / Volume of solution (in liters)

Molarity = 0.01418 mol HCl / 0.02590 L = 0.548 M

So, the molarity of the hydrochloric acid solution is 0.548 M.

I hope this helps! Let me know if you have any further questions.