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Solving Acid-Base Titration Problems

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We're going to work on solving acid base titration problems. To begin with we review the mole ratio. Mole ratios are going to be used to determine the number of moles needed to neutralize a solution. Neutralization is a key word here. So when we combine an acid which has the generalized formula of HA (so we can say this is our acid) with a base with a generalized formula of BOH (here's a base). An acid plus a base combine together to create a salt; where the cation came from the base and the anion came from the acid. Salt plus water. Well where did the water come from? The water came from hydrogen and hydroxide reacting together. So what's important is that we have hydrogen ions coming off the acid. The acid donates the hydrogen ion and the hydroxide ion is coming off the base. And when those species get together - wooo - we make water. And when we've neutralized all of the hydrogen ions with all the hydroxide ions necessary to create water, and not have any extra hydrogens or hydroxides we're done. Okay, and this is what we'll be solving for in the titration problem. But these ratios are going to be really important for us in figuring out how many hydrogens do we need, how many hydroxides do we need. So the ratios are going to be important so we need a correct balance equation with the correct salt product in order to solve the problem. Additionally, we're going to go back to using the mole map. So in the mole map that we started with in the beginning of the year we had 3 arms - the volume of the gas, mass, and representative particles. And we added into this an extension - molarity - or we could say 'concentration'. And we said that molarity is equal to moles of solute over liters of solution. So I'm going to add molarity to both sides of my mole map here. So we may not necessarily be solving for molarity but if we're working with concentrations we want to use this molarity equation. And we said we could connect the double mole map by using the mole ratios from the balanced chemical equation. So when we solve problems involving acid base titrations we're typically going to start out right over here. So we're going to start out with a known amount of a known concentration. Okay, then we're going to have to go through several steps. We start with what we know, we'll use the molarity equation in some form to get this to moles, then we'll go from moles of our known, to moles of our unknown using the mole ratio. And then we'll go from moles of our unknown to molarity or volume in the next step. So three main steps. They all stem off of the first step which is actually the balanced chemical equation. Now typically this is the hardest part for students so you want to make sure that you remember when you're balancing chemical equations you first have to balance each individual substance your reactants and your products species. So don't forget to look for charges and have you written the right subscripts and is a subscript supposed to be there? Once you have a balanced chemical equation, we're going to follow the mole map to help us out. We'll first identify the number of malls of known in our standard solution and since we're typically going to be working with molarity we know that molarity equals moles over volume, so what we want to do is rearrange this and we know that moles will be equal to molarity times volume. Okay, molarity times liters. Then the next step is going to be to convert to moles of our unknown using the mole ratio. And then finally we'll solve for our unknown. And when I say we're going to solve for our unknown, we could be solving for molarity, or we could solve for volume; we could solve for liters. Once again molarity equals moles over liters. If I'm solving for liters, then liters will be equal to moles over molarity. Hey, so being able to rearrange equations is important. Let's take a look at a sample problem, here we go. How many milliliters of .45 molar hydrochloric acid must be addd to 25 milliliters of 1.00 mole potassium hydroxide to make a neutral solution? So step number one is write the balanced equation. Hydrochloric acid plus potassium hydroxide. Alright, rearrange, we're going to form a salt - potassium chloride and water, H20. And I can look and see that I have one potassium on each side, one chlorine on each side, 2 hydrogens on each side, one hydroxide on each side - so this one is nice and balanced. Now when we did our stoichiometry problems earlier in the year, I said to use the equation to help you organize the information. So let's do that again. So underneath hydrochloric acid I'm going to write the concentration which is .45 molar. Hey, and I don't know what volume. That's what we're trying to solve for. Under potassium hydroxide I'd have a 1.00 molar solution, and I have 25 milliliters. So this allows me to see all of the information in a simple format. so now I can get into solving the mathematics, the stoichiometry - the titration kind of stoichiometry for this problem. So first I want to find moles of my known solution, well what is my known solution? My known solution is what I know the most about, so in this case I know the most about potassium hydroxide - so that's going to be my known solution. Okay, reviewing once again if molarity equal moles over volume then moles is equal to molarity times volume. So that's what I'm going to do here alright? I have 1.00 molar times .025 liters okay, remember we want to use liters, so that gives me .025 moles of my potassium hydroxide solution. Hey, so I found moles of my known. Now I'm going to use the mole ratio to find moles of the unknown. So this time I have .25 -oops sorry about that - I have .025 moles of potassium hydroxide my mole ratio is one to one but let's write it out for ourselves because not every problem is going to be on to one so get in the habit of doing this now. Moles of what I'm going to on top, moles of what I'm going to on the bottom remember the ones come from the balanced chemical equation and so this time I know that I still have .0 25 moles of HCl. HCl is my unknown, my unknown is HCl. That's what I"m trying to solve for. In this particular problem I'm being asked to solve for volume. Once again, if molarity equals moles over liters rearranging this equation I get liters - or volume - equals moles over molarity. So, I found that I have .025 moles HCl .45 molar is the concentration the problem gave me, Solving this, I get an answer of .056 liters and if you move the decimal, one, two, three places... then you get the answer of 56 milliliters. So take a moment to look at all the steps. We started with the balanced chemical equation - check! We found the moles of the known solution - check! We used the mole ratio to find moles of the unknown - check! And then we solved for the unknown - and in this case we were solving for volume. So let's see another problem. In this problem we're being asked to solve for concentration. Notice the steps are exactly the same. We'll just have a rearrangement of the molarity equation in number four, which is actually a little bit easier because for solving for molarity and not volume this time. So once again start with the balanced chemical equation You've got a lot of different species to deal with here, so it can be challenging. Calcium hydroxide (CaOH2) plus acetic acid (HC2H3O2) goes to salt - calcium acetate balanced for all the species and water going back through and balancing them Sorry about that. I'm going to need 2 acetates and I'm going to need 2 waters So, balance chemical equation. Once again organize the information that was given to you in the problem. So if I take a look at the problem I'm told I have 24.6 milliliters and I want to find molarity. And then I have 14.2 milliliters and the known molarity are ready. So remember, the moles of known solution is the one that you know the most information about. So I know the most information about acedic acid. So for acedic acid, I'm going to find moles of my known, so molarity equals moles over liters so moles equals molarity time liters In this case I have .0140 molar times .0142 liters (remember I've converted milliliters to liters) and I get a complicated looking number 1.988 times 10 to the -4th moles. Very, very small number of moles. Now I'm going to use the mole ratio to find moles of my unknown 1.988 times 10 to the -4th moles - this is of the acedic acid - Alright, so H2 (long pause) HC2H3O2 (bell ringing in background) You can tell I'm in my classroom doing this. 1 mole of my unknown which is what I'm solving for on top 2 moles from the balanced chemical equation of acedic acid. And I get 9.94 times 10 to the -5th moles of my calcium hydroxide. Now I must solve for the unknown. In this problem I want to know moles. Moles is just - Molaritty. Molarity is just volume so I have moles.... volume comes all the way from up here Alright, so I've got to cover that into volume in liters so .0246 liters. And my molarity here is a small molarity 4.04 times 10 to the -3rd molar. So I know it's a lot of information to look through but you've seen this before we've done these types of problems before using the double mole map; it's just using the molarity component as part of the double mole map. Write a balanced chemical equation. whichever one you know the most information about find the moles that substance. Then use mole ratio to find that your other substance. And then solve for the unknown; whether you're solving for molarity, solving for volume, is just going to use the same equation. So keep all those steps in mind as you're solving titration problems

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Duration: 13 minutes and 18 seconds
Language: English
License: Dotsub - Standard License
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Views: 73
Posted by: christineward on Sep 12, 2015

Solving Acid-Base Titration Problems

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