Titrations Calculations part 2

Okay, titration calculations, part 2. Calculate the volume of each solution that contains the following numbers of moles. So we want to know. We're given the number of moles, we're given the molarity in moles per liter. So we have to find the volume. So this is a. Well, let's go into here. Let's just put our original first. Moles equals molarity times volume. Therefore, what does volume equal? Well, volume equals moles over molarity. It's as simple as that. There's our base equation. So the first thing we've got to do, then, we've got to say how many, what's the volume of this particular system? So V equals. And we've got the number of moles, 0.005. We've got the molarity, 0.1. Divide anything by 0.1 is multiplied by 10. So it's 0.0 liters DM cubed. DM cubed. It doesn't say many particular units. We have to put them in. So we can leave it in. DM cubed. 0.05 is 50 cm cubed. Incidentally, the next one, it's going to be the same system again, isn't it? So we've got volume. Volume equals. We want the number of moles 1 times 10 to the minus 5. Okay, well, let's leave it as 1 times 10 to the minus 5. We divide that by. We divide that by the. By the molarity, which is 0.01 divided by 100 is the same as multiplying. So divided by 0.1 is the same as multiplying by 100. So we'll just take our two decimal places. What times 10 to the minus 5 becomes 10 times 10 to the minus 3. So it's 1 times 10 to the minus 3. Now, we just put that in there. 1 times 10 to the minus 3 is 0.001. Of course, we want to put it in here. V equals 0.001. And again, it's liters DM cubed. And into a nice pretty blue color. Question 4. Calculate the concentration of a solution of sodium hydroxide, given that 25 cm cubed required 18.8 cm cubed of sulfuric acid, presumably for neutralization. We seem to have missed the words for neutralization out or for complete reaction. Well, we can see from our equation here that one mole of sulfuric acid is equivalent to two moles of sodium hydroxide. All right? Okay. So one mole. So however many moles of sodium hydroxide or sulfuric acid we've got. We must have twice the number of moles. So we work out the sulfuric acid moles. Moles of sulfuric acid equals molarity times volume. Molarity 0.05. The volume is 18.8 cm. Kubed 0.0188. Okay, let's bring in a calculator for this. It's a bit of mental arithmetic. Otherwise 0.05 times 0.0188 and that comes to three zeros, 0.0094 moles. Then we have to use our equivalence in the equation here. That one mole is two moles of sodium hydroxide. So that. So then we have moles of sodium hydroxide. Moles. Sodium hydroxide equals this number of moles times 2, 9, 4 times 2. And we're going to bring in our calculation for this as we've already got the number on there multiplied by 200188. 0.00188 moles of sodium hydroxide. Now we must apply the relationship to get. Well, it's the same relationship, isn't it? It's the moles of sodium hydroxide equals molarity times volume. Only this time we've got the moles and we want the molarity. So we could say molarity equals. We rearrange this equation. Molarity is to protect the volume to the other side, so it's moles over volume. In this particular case it's 0.00188, which the moles we divide by the volume and we're told it's 25cm cubed 0.025. We bring in our calculator. Again, divide by 0.025 and that gives us 0.0752 0.0752 and it's moles per litre. Moles per litre represents as a capital M, Calculate what volume of this potassium hydroxide is required to neutralize 25cm cubed of that amount of nitric acid. So we can work out the amount of nitric acid we have first of all. So the moles of acid. Moles of acid equals molarity times volume. MV. Let's keep reinforcing this molarity. 0.015. The volume 25 centimeter cubed 0.025. And we'll bring a calculator in here for that 0.015 times 0.025. That comes with three zeros. Three seven five. 0.000375 moles of acid. Well, you can See from the equation, hopefully you can see from the equation that one mole of acid is equivalent to one mole of base. So this is also equal to the mole of base. Now we can use our. This is our fundamental equations. Moles is mb. Well, in this case it's moles of base is mv. We want the volume. So V equals mole over. Whoa. Mole over m. And this moles is 0.00375. Divided by the molarity in the autolent is 0.05 moles. 0.05 moles per liter. So we divide by 0.05 and that comes to 0.0075. 0.0075 its volume. So it's DM cubed. Now of course that's 7.5 cm cubed. Is that a realistic answer? You can always ask yourself, is it a realistic answer? Well, what are we talking about? 50cm of potassium hydroxy, about 0.015. So about 50. That's 0.05 molarity. And the molarity is a lot less. Yes you do. It's a realistic ballpark figure. Question 6. A 250 centimeter cubed solution sodium hydroxide was prepared. 25 centimeters cubed. This solution required 28.2 centimeters cubed of zero. So we can work out the actual amount of hydrochloric acid in the titration. Let's have a look at this then. Okay, so let's change this to an arrow while we're at it. Now we can see it's a one to one reaction. There's one of those hydrochloric acid. So it serves the same moles of hydrochloric acid as sodium hydroxide. So let's work out the hydrochloric acid first of all. Then 28 centimeter cubed. So the moles mole mole molecule equals 0.0282 multiplied by the molarity 0.1. Well, we can see the move the decimal place 0.00282 moles of acid. And this is equal to exactly the same moles of bas. So this only moles in 25cm cubed of the solution. But the original solution has twice as big. So therefore in 250mil of solution there are 0.00282 times 2,10 mol. And this is equal to 0.0282 mole. Well, we know how many moles of sodium hydroxide there are. So now it's just a question of working out the mass. Or mass equals moles times relative mass. Let's be consistent. Use the same symbols, moles times relative mass. The relative mass of sodium hydroxide is 40. So it's 0.0282 times 40. And this is going to give bring in our calculator. Don't know what this value is for. So we take that off. 0.0282 times 40 is 1.128 grams. Right. So that concludes the second part of the titration solutions.