- Solutions 16

Indicate the concentration of each ion or molecule present in the following solutions: (d) a mixture of 45.0 mL of 0.272 M NaCl and 65.0 mL of 0.0247 M (NH_{4})_{2}CO_{3}. Assume that the volumes are additive.

Hey guys, so this question's a little bit different from the ones before it. Here we're dealing with a mixture of two compounds. So they tell us that they have, 45 mLs of 0.272 molar of sodium chloride. Also we have 65 mLs of 0.0247 molar ammonium carbonate. Alright, so, what we need to first realise is, remember the word of means multiply. So when the word of is in between two numbers, it means multiply. And remember that molarity equals moles over litres, and realise when you multiply both sides by litres that moles equals litres times molarity. So we have to change these millilitres given to us into litres. So we have 45 mLs for every one milli- it's ten to the negative three litres. Remember molarity really means that number in moles divided by one litre, so that's going to give me the moles of sodium chloride. And what we can realise here is that the moles will be at this point 0.01224 moles of sodium chloride. Those moles I can change into the moles of sodium and chloride individually. So one mole of sodium chloride here, is one mole of sodium, so it's the same number, and if you did it with chloride ion you would see it's the same number as well. Okay, so both of them have the same moles.

Now we need to find their concentration, but here's the thing, concentration is the same thing as molarity. Molarity equals moles over litres. Our original volume was 45, but remember, we took 45 mLs of this compound and mixed it with 65 mLs of this compound. So we have a new volume. So my new volume is the 45 mLs plus the 65 mLs. So that's a 110 mLs. That I need to change into litres, so my new volume of my solution, when I convert it into litres, equals 0.110 litres. So to find the concentration of sodium and of chloride individually, just divide them both over that total volume. So we're going to take those mole that we took for each one and divide them by this volume here, which means that the molarity of sodium is this, and the molarity of chloride is this as well.

So we found out the molarities of those first two ions, now we have to do it for ammonium ion and carbonate ion. So here we're going to change out 65 mLs into litres as well. So one mL is ten to the negative three litres. Remember, molarity really means moles over litres. So here what we're going to get, is we're going to get 0.001606 moles of the entire thing. If we wanted to we could just divide this by the volume and get the molarity of the entire compound first, and then we could figure out the molarities of each individual ion if we wanted. So I'm going to do this one a little bit different from the first one, so you can see variety. So here's the molarity of the entire compound. Now let's find out the molarities of each ion within this entire compound. So remember here this molarity here that we just found, really means moles over litres. So we're going to say here for every one mole of the entire compound, look how many moles of ammonia we have, we have two moles of ammonia. So that's going to give me 0.0292 molar of an ammonium ion, and we do the same thing for the carbonate ion. So we have the molarity of the entire compound here, divided by one litre, for every one mole of the entire compound we have one mole of carbonate, there's only one mole of carbonate in the formula. So that gives me 0.146 molar of carbonate ion.

So that's the approach, two different approaches, which would still get us the same answer at the end. You can either find the molarities of each ion individually, or you can find the molarities of the entire compound then change the molarity of the entire compound into the molarity of the ions. Either way they will still get the same answer