26.0246 g of sodium carbonate is dissolved in water and the solution volume adjusted to
250.0 ml in a volumetric flask. A 25.00 ml sample of an unknown hydrochloric acid solution required 32.06 ml of the sodium carbonate solution for complete reaction. What was the molarity of the hydrochloric acid solution?
My Work:
2HCl+Na2CO3-->2NaCla+H2O+CO2
Step 1: 26.0246g/105.99g (Molar Mass of Na2CO3)=.2455 mols in 250ml
Step 2: .2455mols*(.03206L/.250L)=.03148 mols (in 32.06 ml)
Step 3: .03148*2 (because of the 2:1 mole ratio)=.06296 mol HCL in 25 ml
Step 4: Molarity=mols/liter=.06296mol/.025L=2.519 M
Am I on the right track? Thanks
I didn't check your work but I came out with the same value at the end.
Yes, you are on the right track. Your calculations are correct and your steps show a clear understanding of how to calculate the molarity of the hydrochloric acid solution.
Just to recap, you have determined that there are 0.2455 moles of sodium carbonate in the 250 mL solution. Then, you used the stoichiometry of the balanced equation to determine that there are 0.03148 moles of HCl in the 32.06 mL of the sodium carbonate solution.
Finally, by comparing the number of moles of HCl to the volume of HCl solution used (25 mL), you calculated a molarity of 2.519 M. This means that there are 2.519 moles of HCl per liter of solution.
Overall, your work is correct and you have successfully determined the molarity of the hydrochloric acid solution. Well done!