Hi there. In this problem, they are neutralizing some sulfuric acid with sodium hydroxide. So the first someone to do is write a balanced equation for that reaction. So what I have is H two s 04 thank you. Reacts with two n a. O. H to produce and a two s 04 and to H 20 This is a neutralization reaction where an acid reacts with a base to make a salt and water and to try to determine the mole Arat e of this h two s 04 we use 100 mL for dividing that by 1000.1 leaders. The mole arat E is what we're trying to find. So they add some sodium hydroxide to that to neutralize it, but end up adding too much sodium hydroxide. So now the solution is basic. So to correct that situation, they add some HCL to the endpoint or until we have perfect neutralization. So I also want to write the reaction for that. HCL reacts with n a O. H. And that produces in a C l and H 20 We know that we use 13.21 mL of the HCL to neutralize that excess. N a o h. I'm going to convert that toe leaders by dividing by 1000. So that gives me 0.0 13 to 1. Leaders of this and it's molar ITI is 0.103 This is all information given to us in the problem. The sodium hydroxide that was used 50 mL were used of that which, divided by 1000 gives us 10000.5 leaders and its concentration this point to 13 Moeller. Right? So the first thing I want to do is constantly is calculate how many moles of H. C l were used and how many moles of any a wage were used. So let me go ahead and do that. Starting with the HCL, I have 0.0 1321 Leaders of that have that solution. And if I calculated by the molar ity polarities of course moles her leader that will give me zero point 001 3601 rolls of hcl. All right. And then for the O. H. We have 50 mL or 0.5 leaders of the O. H, multiplying by the way, which is more clarity. Yeah, okay, gives me zero point 010 65 moles of anyway, H, this is how much anyway h is available. We can see that it is definitely mawr than the moles of HCL. So if we want to figure out how much anyhow wage was used by the HCL, we look back to the balanced equation and we see that as a 1 to 1 ratio. So if we used 0.136 So one moles of HCL, we're also going to have to have used 0.1 3601 moles of any ohh to neutralize that. Okay, so that's how much any of which was used to neutralize the HCL. And we can see how much anyway, which was available. So I'm going to subtract those two values to see how much any a wage remains because that is Thea Mount. That would neutralize our H two s 04 So I'm just going to take these two numbers and subtract 0.1065 moles minus 0.0 13 601 moles of anyway, H, remember this was the amount used. Looking in this first amount was the amount of any a wage available. The difference gives me zero 0.9 to 899 I will around a significant figures here in a few minutes when I get to the end of the problem moles of any of which. So let's think about this is this is the amount of any wage that was used to neutralize the H two s 04 Let's go ahead and do a little stuck geometry with that. If I have 0.0 9 to 8 99 moles of any O. H. Looking back to my first equation, I see that the mole ratio his two moles of any wage are needed for every one mole of the H two s 04 Doing this math gives me zero point 00 464 moles of H two s 04 That's the amount of H two s 04 that was neutralized by this base. So that's how many moles of H two s 04 were in this solution. We know that morality is defined as moles divided by leaders. So we now have the molds of H two s 04 Right? And we know from the problem that the amount we were testing was 100 mL or 1000.100 leaders. So taking moles divided by leaders is going to give us our modularity. Our polarity then is going to be four 0.64 times 10 to the negative, too. Final answer. This is theory, Jinnah. Concentration of our H two s 04 solution. All right. Thank you so much for watching, Okay?