A 1.000 g sample of Na 2 CO 3 (MW: 105.99 g/mol) is dissolved in enough water to make 200.0 mL of solution. A 25.00 mL aliquot required 32.18 mL of HCl to completely neutralize it. What is the molar concentration of HCl? Na 2 CO 3 (aq) + 2 HCl (aq) → 2 KCl (aq) + H 2 O(l) + CO 2 (g)

We can consider this titration question a bit of a cool down based on the other one that we just saw in terms of titrations and back titrations. But we still have to be very careful in our approach to answer the question. Here it says, A 1 gram sample of sodium carbonate is dissolved in enough water to make 200 ml of solution. We are told that a 25 ml aliquot required 32.18 ml of hydrochloric acid to completely neutralize it. What is the molar concentration of hydrochloric acid? Now remember, concentration means molarity. We're being asked to find the molarity of HCl. That would be moles of HCl divided by liters of HCl solution. Technically, we already know a portion of this because we know the amount of ml's of HCl solution that was used. So change those ml's into liters. So we plug that here on the bottom. And all we have to do now is figure out the moles of HCl. Here we're told that we have 1 gram of sodium carbonate. We're given the molar mass of sodium carbonate. So we're going to plug that right in. And that's for every 1 mole of sodium carbonate. So when we do that, we're going to get 0.009435 moles of sodium carbonate. Now, we're going to say here, they're telling us that we dissolved, this amount of Na, sodium carbonate to 200 ml of solution. So we're going to say with these moles here, we have connected to it 0.200 liters and we're only taking 25 milliliters of it. So we have the way we'd set this up to make it easier to understand is we could just say m1v1 equals m2v2 where m here doesn't have to represent molarity. It could represent moles. So we're taking 25 ml's of this, which is 0.025 liters, and we're diluting it to 200 ml's, so that's 0.200 liters. Divide both sides here by 0.200 liters, and that's going to give you my new moles of sodium carbonate. So that gives me 0.00-1179 moles of sodium carbonate. Now that I know the true amount of sodium carbonate after the dilution has happened, we're going to now do a mole to mole comparison and say that for every 1 mole of this, there are 2 moles of this. So that's going to give me 0.00-2359 moles of HCl. Take that and plug it up here. And so that equals 0.073297 Molar HCl. And if we reduce that or round it to 3 sig figs, it comes out to 0.07330molar hydrochloric acid solution. So that would be the concentration of my acid. Now again, this is just a typical titration question. We're not looking at any excess of anything to figure out some type of back titration type of scenario. Here, we just have to figure out the tricky part was dealing with the dilution of my sodium carbonate solution. But setting up m 1v1 equals m 2v2 can help you determine what the new moles of sodium carbonate would be. Once you figure that out, you can just do a simple stoichiometric relationship where we compare the moles of sodium carbonate to the moles of hydrochloric acid to figure out the moles of HCl so that we can get our molarity at the end. Guys, remember, take to heart some of the techniques we learned in terms of regular titrations as well as back titrations.