Ah well, I tried to work backwards from the answer and I'm no where close. So it's easy to say that I haven't done anything right.

Yeah, so like, answers a crutch - you don't really learn if relying on answers or worked solutions. Think of doing your QCE like you were running a race. Let's say you go out all the time and practice. Every day. But all you ever do is walk. Then, it comes time to do that race - your body is used to being outside, it's used to moving a lot, but the moment you start to run, it has no idea what to do, because it's too used to walking. Sure, you'll probably do better than if you didn't do all that walking, but will you do as well as you possible can? Always try just doing a question first - even if you don't know where to start, just try something. You'll learn a lot more that way

As for this one, we have a lot of numbers that are potentially just there to throw us off. So let's break everything up into steps:

1. Sodium carbonate solution was made. [Na

_{2}CO

_{3}] is now known. Numbers were 5.267g of sodium carbonate in 250mL of water.

2. Sodium carbonate was titrated with unknown HCl. [HCl] is now known. Numbers were 10 mL of sodium carbonate solution with 21.3 mL of hydrochloric acid solution.

3. HCl was titrated with unknown Ba(OH)

_{2}. [Ba(OH)

_{2}] is now known. Numbers were 25 mL of barium hydroxide and 27.1 mL of hydrochloric acid.

So, let's do a together. So, if we look at our steps, we need to look at step 2. The easiest way to do this is to start with an equation we KNOW we have to use, then figure out what we do and don't have in that equation, and work backwards. So, to figure out the concentration of HCl, we use the equation:

We know V, but not n. Okay, so we need to find n - we can use the fact that 10mL of the carbonate solution was used to react with the HCl. Okay, since they react by the chemical equation:

We can use the mathematical equation:

Okay, so we need the amount of mol of sodium carbonate. Well, to find that, we HAVE to use the information in step 2. Remember - if you want the mol from a step, you ALWAYS have to use the mol from the same step. So that would mean the mol from the 10mL, so we use the equation:

But we don't know the concentration for sodium carbonate!! So, the question is, can we use the same concentration from step 1 in step 2? The answer is yes - you can never use information from a later step, but you can use information from a previous step if it does not change. The concentration of sodium carbonate does not change from step 1 to step 2, so we can use it. This means we use the equation:

I'm just gonna call those n1 and V1 to make my life easier. Okay, so we know the volume is 250mL - but we don't know n! So, we have to find n. Ah, but we do know that sodium carbonate was weighed, so we can use the equation:

And finally, we have all of those numbers! So, we calculate n1:

And we can finally go through all of our previous equations:

which gives us the final molarity of HCl - 0.0467.

So I'm sure this felt like a lot of maths to do all at once, but remember - all we did is figure out what equation we had to use, and then just went backwards one at a time - which should be a little less daunting if you go through this process yourself. Why don't you now try b and c, and see if you can do those yourself?