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
1 answer
I didn't check your work but I came out with the same value at the end.
1 answer