A 0.5895-g sample of impure magnesium hydroxide is dissolved in 100.0 mL of 0.2050 M HCl solution. The excess acid then needs 19.85 mL of 0.1020 M NaOH for neutralization. Calculate the percentage by mass of magnesium hydroxide in the sample, assuming that it is the only substance reacting with the HCl solution.

Video transcript

Hi everyone for this problem, we're told a sample of barium hydroxide that is contaminated with some impurities has a mass of 1. g. This sample was dissolved in 200 mL of 2000. molar hydrochloric acid solution. The excess hydrochloric acid was tight traded with . Molar sodium hydroxide solution and 28. ml of the solution was used. We need to determine the mass percent. So let's make a note there. The mass percent of barium hydroxide in the contaminated sample. No assume that only the barium hydroxide in the sample reacted with the hydrochloric acid solution. Okay, so we want the mass percent of barium hydroxide. So let's recall what exactly that means. So mass percent of barbarian hydroxide is going to equal our mass of barium hydroxide and grams over our total mass and grams of our sample. Okay, so in the problem they tell us that our total mass of the sample Is 1.316g. So we know that part. So we know that it's 1.316g. But we need to calculate the mass of barium hydroxide and we need to write out a reaction for this. Okay, so our reaction is going to be are barium hydroxide reacts with our base to produce a salt and water. This is our balanced equation. And the first thing that we need to do is calculate the total moles of hydrochloric acid. That is the first part. So they tell us that the sample was dissolved in 200 mL of 2000.2250 molar hydrochloric acid solution. So step one is going to be to calculate our total moles of hydrochloric acid. Okay, This is what's going to lead us to calculating are massive bearing hydroxide. So let's start here. So what we're told in the problem is that we have 200 ml of hydrochloric acid and we need to go from volume of hydrochloric acid, two moles of hydrochloric acid. So we first need to convert this volume from middle leaders to leaders and one male leader We have 10 to the negative. Three leaders. Our male leaders cancel. And now we can use Armel arat e. That was given to go from volume to moles. So Armel aren? T remember malaria is moles over leader. So we're told we have 0. 50 moles of hydrochloric acid for every one liter of solution. Okay, so now we're left with units of moles of hydrochloric acid, which is what we want. And when we do this calculation we get 0.045 moles of hydrochloric acid. Then they tell us that the excess hydrochloric acid was titrate, id with molar solution of sodium hydroxide and 28.5 ml of hydrochloric acid was used. So now we need to calculate the moles of excess hydrochloric acid. Okay, so moles of access hydrochloric acid. So let's write the reaction for what they tell us. They tell us that our excess hydrochloric acid was titrate, id with sodium hydroxide. So our reaction is going to be hydrochloric acid plus sodium hydroxide. This acid and base is going to produce a salt and water. That salt is sodium chloride plus water. Okay, so now we need to go from volume of hydrochloric acid to moles of excess hydrochloric acid. Okay, so let's go ahead and do that. We have we're told we use 28.5 ml of solution. So that's 28.15 ml of hydrochloric acid. And we need to go from volume of hydrochloric acid, two moles of excess hydrochloric acid. So from milliliters to leaders, we have one male leader is equal to 10 to the negative. Three leaders, we can use the mill arat e that we're told for our sodium hydroxide, which is 0. moles of sodium hydroxide for every one liter of solution. So let's just make sure our units are canceling properly. Our middle leaders cancel our leaders canceled. And now we're in moles of sodium hydroxide and we want to go to moles of excess hydrochloric acid. So we can use our multiple ratio here. Looking at our problem, we have one mole of acid for every one mole of base. So we can use that to go from one reacting to the other. Okay, so for every one mole of sodium hydroxide, we have one more of hydrochloric acid. Okay, so now our moles of sodium hydroxide cancel. And we're left with moles of hydrochloric acid and this is going to give us our molds of excess. So when we calculate this out, we get three .16406 times to the -3 moles of excess hydrochloric acid. So now that we know our molds are total moles of hydrochloric acid and our molds of excess hydrochloric acid, we can calculate the Actual moles of hydrochloric acid reacted by subtracting these two values. Okay, so step three is going to be figuring out the moles of hydrochloric acid. Actually that reacted. Okay, so we're going to subtract these two values. We have 0.45 moles of hydrochloric acid minus our excess hydrochloric acid 3.16406 times 10 to the negative three moles of excess. And that gives us 0. 84 moles of hydrochloric acid reacted. And the reason that we did this is because we're told that we had hydrochloric acid and then excess hydrochloric acid. So we needed to calculate our actual moles of hydrochloric acid reacted, which is this right here. So now that we know our actual moles of hydrochloric acid that reacted, We can now figure out the mass of barium hydroxide Using our moles of hydrochloric acid reacted. So step four is going to be to get the mass of barium hydroxide using are moles of hydrochloric acid reacted as our starting point. Okay, so now we know that we have 0. moles of hydrochloric acid reacted. And Going back up to our reaction at the top we have a multiple ratio of 1-2. So for one mole of barium hydroxide We have two moles. Make that a little clear. We have two moles of hydrochloric acid. Okay, so our moles of hydrochloric acid cancel. And we have moles of barium hydroxide and we want the mass of barium hydroxide and grams. So what we can do now is use the molar mass of barium hydroxide to go from moles to grams and one more of barium hydroxide. Our Molar mass is 58. 21 g of barium hydroxide. Okay, so our moles of barium hydroxide cancel. And we're left with grams of barium hydroxide, which is what we want. So we get 1.22 zero g of barium hydroxide. So now we figured out our missing piece to our mass percent equation. So let's scroll back up and plug that in. So our mass of barium hydroxide and grams is 1.220g. And we're going to multiply this by 100% because we want a percent. Okay, so once we do that, we get a final answer of 92.71%. And this is our final answer. This is the mass percent of barium hydroxide in the contaminated sample. That's the end of this problem. I hope this was helpful.