What volume of 0.01060 molar HBr is required to neutralize 25 millilitres of 0.01580 molar barium hydroxide? So here we're just looking at a stoichiometry problem, because we're given amounts of one substance and asked to find some unknown quantity for the other. This happens to be an acid based reaction, and it actually is a titration problem, because we're talking about neutralization. But we still take the same approach that we do for any stoichiometry problem. A couple of things that we need to note, is one, we're not given an equation, so we're going to have to first write a balanced equation before we can look at anything related to stoichiometry. The other thing to note is to remember what molarity means, that molarity equals moles over liters. And so, instead of thinking of this, say, as 0.01580 molarity or molar, think of it as moles over liters. And that can help you see how you're supposed to use it in the calculation. So, the first thing we want to do is write that balanced equation. So, I know I have two reactants, HBr and barium hydroxide. I know when I look at my products for an acid base reaction, I'm going to use a salt and water. So, I can go ahead and write in the water. And now I have to figure out what the formula is going to be for the salt. Note that my cation is going to be barium and my anion is going to be bromine. So bromine, when I looked at my ions, I know that barium has a two plus charge and bromine has a minus one charge, so if I crisscross down, I see I'm going to have BaBr2 for my formula for the salt. Now, I need to balance the chemical equation. And I'm going to actually have to put a 2 here in front of HBr as well as a 2 in front of the water. So now what I have, is a balanced chemical equation. If you're unsure how I got to that, make sure you stop here and actually work through the balancing of this chemical reaction before you proceed. Now, I want to look at what information I know and what I'm trying to find. So in the problem, it's given me the volume and molarity of barium hydroxide and just the molarity of HBr. Well, with just the molarity of HBr, I can't get to moles. And remember, that's that central part of every stoichiometry calculation that I have to go through Moles. Since I don't, can't or I don't have the information to get to moles of HBr, that tells me I must be starting with the information about the Ba(OH)2. So I start with my 25 millilitres of Ba(OH)2. And I'm going to convert that to litres, so 1000 millilitres in 1 litre. Now I'm going to use my molarity of Ba(OH)2 and I'm going to rewrite the units as I put this into my equation as moles of Ba(OH)2 per litre. So now, what I see is that if I start cancelling, my milliliters cancel with milliliters, litres cancel with litres. At this point, I'm left with moles of barium hydroxide. Now, I'm going to do just like I do in any other stoichiometry problem. I'm going to use the coefficients from our balance chemical equation in order to get from the barium hydroxide to the HBr. So, I see for barium hydroxide I have 1 mol of the barium hydroxide per 2 mols of HBr. And this is the critical step in these calculations and the one that students miss most frequently. So, make sure you always put this molar ratio, no matter what type of stoichiometry problem you're looking at, you always have to put that mole ratio in there. Now if I look, I get my moles of barium hydroxide cancel with one another. Now I'm left with moles of HBr. But there I can actually figure out the volume of HBr, because I know the concentration. I'm also going to rewrite that in terms of mols over litres, but this time I'm going to have to put the moles on the bottom, because now I want the moles to cancel out, so my moles of HBr to cancel with moles. And now I can do the calculation, and when I do that what I end up with is that my volume of my HBr is going to be 0.07453 litres or I can write that as 74.53 milliliters. Again it's about making sure that our units cancel out correctly.