Titrations Calculations part 3

Titration calculations. Part 3. What volume? 5 molar hydrochloric acid is required to neutralize 20 kg. Wow. Of calcium carbonate. Okay, first thing we do is we work out the moles of calcium carbonate. Moles of. Spell it right. Calcium carbonate. What's the mass divided by the relative mass? Well, the mass is 20,000 divided by the relative mass, which is a hundred. It's actually quite easy to work out. You just cross two naughts off and it comes to 200 moles. Right. Now we. Now we apply moles equals MV. This is our equation for solutions, don't forget. And what do we want? We want the volume. So we want V equals moles over M. The moles we've got is 200. M is 5. And so we're left with 5 into 200 is 40. So it's 40 liters. 40 DM cubed. Bucket chemistry. All right, next question. Question eight. We're given the number of grams of a monoprotic acid. That means it's got one hydrogen. Mono means one protic. Means hydrogen or proton was dissolved in water and the solution made up to 250. We took a tenth of this 25 and titrated it with solution. And this is how much of the solution we required. So the information we can get, first of all is the number of moles of sodium hydroxide. Right? So we go moles. Moles of sodium hydroxide equals molarity times volume MV, and it's 0.095 times the volume, which is 0.0465. Not a particularly nice calculation. Let's bring this one in. Calculated for this point zero times 0.0465. And that comes to 0044-1750-0004-4175 moles. Now we're told it's monoprotic. And that means that will have the same number of moles of acid because there's one OH group in sodium hydroxide. And a monoprotic acid will have formula of HA. So moles of HA will be the same 0.00. Moles. Now, this is in 25cm cubed. Of course, the original solution had 250. It's 10 times as big. So the moles of HA in 250mil equals zero point. Just take one naught off. Multiplied by 1025 into 250 is a factor of 10. This is the number of moles. Now we're given the mass and we know the relationship between moles and mass. Relative mass, moles Equals mass over relative mass. We want the relative mass. So we take that, become the subject of the equation. Relative mass equals. Take it to the top. It equals the mass over the moles. Mass over moles equals. Well, the mass we've got is 3.88. The moles we've got is 0 point. All of this. So this is a calculator. Let's bring our calculator in. And remember we've got to multiply that by 10. We haven't done that just to get the correct number of decimal places. And then in fact, if we take 1 over X, then in other words the reciprocal of that. Then divide. What have I done? That's the reciprocal of whatever. Then multiply that by 3.88. There's my calculation. And the answer comes to 87.8. Relative mass. So it has no units. I wonder what it is. I have no idea. 87.8. We take one hydrogen off it and you get 86.8. It's CLO3. What would that come to? Just out of curiosity, 48 +35.5 come to? 83.5. No, not quite. Doesn't quite make it. Never mind. Okay, on to the Next question. Question 9. What are we given? A sample of ethandioic acid crystals. They contain two hydrogens to diprotic acid or dibasic acid dissolved in water to make up to 250cm cubed. One mole of the acid reacts with two. There's our information. We need that in a titration. 25 centimeter cubed of this solution of acid reacts exactly with 15 of 0.16 moles per liter sodium hydroxide. So the first thing we can do is we can calculate the moles of sodium hydroxide. Moles of NaOH equals molarity times volume. Remember, always moles is molarity times volume. For solutions. 0.160 times 0.0. 156. What does that come to? It's 0.16 times 0.0156 002496. 0.002496 moles. Now, we're told that one mole of acid reacts with two moles of sodium hydroxide. So moles of acid will be moles of acid. I'm going to call it h not ha moles of acid because it's actually got two hydrogens. Say 2a. Well, it will be half the number of moles of sodium hydroxide. 0024892496. Sorry. 2496 over 2. So we take our calculator in and we just divide this number by 2001248. But remember this was in 25 centimeter cubed. We're actually interested in having it in 250. So let's multiply it by 10. So moles of AC, moles of acid in 250 mil equals the same copy zero point that times 10. So we change the decimal place one moles. Right. Now we're given the mass of the acid. How do we relate the mass to the moles? Well, the moles equals mass over the relative mass. Relative mass, we're interested in the relative mass. So relative mass make it subject. We take it to the top and the moles to the bottom is mass over moles. All right, well the mass we have, let's put the mass, let's do it all the way through. This time the mass is 1.575. And we're going to divide that by the moles which is 0.01248. And that comes to 126. Now then this gives us the mass of the. I've done this wrong, haven't I? I've used. No, no, no, no, no, no, no. Sorry, I'm jumping ahead of myself. This is the moles of acid only. And the acid only is H2C2O4. So the relative mass of the acid, the relative mass of the acid only without the water equals. Well, H2 2 and 24 is 26, 26 and 64, 2664 is 90. So therefore. So the mass of acid without water is equal to the moles times the relative mass without the water equals 0.01248 times 90. So now we bring our calculator in. Point 01248 times 90. That comes to 1.1232. 1.1232 times 90. Just what am I doing? I've multiplied by 90 grams. That means the actual water. So the mass of the water is the weighed out mass which is 1.575 subtracting the mass of the acid only. 1.1232. Let's take this into the. So minus that plus 1.575 and that comes to zero point. Zero point. I've just put equal. Should be equal. Sorry. Equals zero point. I just lost the calculator. 4518. 4518 grams. So now from the mass of the water we can work out the moles of the water. Moles of water equals 0.4518 over the relative mass which is 18, bring that calculator again divide that by 18 and that comes to 0.0251 0.0251 so now we have a ratio of mass of water to the ratio ratio moles of water of acid to the moles of water equals the moles of acid. Remember where's the moles of acid? Moles of acid in zero. There it is. Copy. To 0.0251. And to get that into a simplest ratio integral ratio, we divide ratio we divide by, divide through by the smallest. So it equals if you divide this by this is the smallest one, divide that by that comes 1 acid 2 and that comes to. That's divided by 0.125 approximately 1248 is going to come to 1 to 2. So in other words, the value of x value of x is 2. So in oh, that's actually N the value of n is 2. So the formula would be. Just replace that N with the two and that is the entire solution.