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In this video, we will learn how to use the results from a titration experiment to calculate unknown properties of a solution.
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Exactly how titration experiments work is not really the purpose of this video.
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Instead, we’re going to look specifically at performing calculations from the results of these titration experiments.
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Remember that during a titration experiment, you add small amounts of a solution with a known concentration to a solution with an unknown concentration.
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The solution of known concentration might be referred to as the titrant or a standard solution.
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The solution whose concentration is unknown might be referred to as a titrand or an analyte.
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During our titration experiment, we continue adding our standard solution until our unknown concentration solution has been neutralized or fully reacted.
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These are all important terms that we’ll come across later when looking at example questions.
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And the key point of getting all this data is that we can then use the amount of titrant added to calculate the unknown concentration.
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So let’s see how we do this.
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To carry out calculations with data obtained from titration experiments, we’re going to need a couple of key equations.
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The first key equation we’ll need is 𝑛 equals 𝑐𝑣, where 𝑛 is the amount in moles, 𝑐 is the concentration in moles per liter, and 𝑣 is the volume in liters.
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Let’s practice using this equation before we move on.
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How many moles of NaCl are there in 1.5 liters of 0.3-molar solution?
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First, we need to remember that molar means moles per liter.
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In the question, we’ve been given the value of 𝑣 and 𝑐, and we’re asked to find the value of 𝑛.
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We start with our equation 𝑛 equals 𝑐𝑣.
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And we can substitute in the values from the question.
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Next, we calculate 0.3 molar multiplied by 1.5 liters, which gives us 0.45 moles.
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We can double check our calculation by looking at the units.
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By doing moles per liter multiplied by liters, the liters cancel.
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And we should be left with moles.
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Since moles are the appropriate units for the value of 𝑛, it looks like we’ve done this correctly.
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Now, we need to use this same equation to perform our titration calculations where we have an unknown concentration.
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So let’s look at an example.
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A 0.05-liter solution of HCl was titrated against a 0.1-molar solution of NaOH.
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The addition of 0.025 liters of NaOH was found to neutralize the HCl.
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What is the concentration of the acid?
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Perhaps one of the trickiest things about titration calculations is working out exactly what you have to do from quite a wordy question.
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One way to approach this is to simply strip out all of the really important information and lay it out in a logical way.
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But we should also try to understand exactly what’s going on here.
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So to do that, let’s try writing out a really simple balanced equation.
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The question tells us that we are titrating HCl with NaOH.
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So these are our reactants, but what do they form?
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This is actually an acid–base titration.
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The question uses the word “neutralize,” which gives us a clue.
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Now that we know this, we can work out what our reactants form.
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Of course, we get a salt, in this case sodium chloride, and water.
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The important thing now is to check that this is balanced.
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You’ll see why this is important a bit later on.
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Now let’s get the important data out of our question.
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Let’s try laying out all of our information in a handy table.
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First of all, we know that we have 0.05 liters of HCl.
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So we can put this in the appropriate volume box.
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Next, we’re told that we have a 0.1-molar solution of NaOH.
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Remember that molar is the same as moles per liter.
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We’re told that we need 0.025 liters of our NaOH to neutralize the acid.
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So again, we can add this to our volume box.
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The question is asking us for the concentration of the acid.
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So we can label this on our table as our goal.
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Now we just need to fill in enough gaps on our table until we manage to fill in the concentration of the acid box.
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Remembering our key equation 𝑛 equals 𝑐𝑣, we can work out the amount of moles of NaOH.
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To calculate the number of moles of sodium hydroxide, we multiply the concentration, 0.1 molar, by the volume, 0.025 liters, which gives us 0.0025 moles.
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Now comes the important part of why titrations are really useful.
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From our balanced equation, we can have a look at our molar ratio.
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We can see that we have a one-to-one ratio of acid to base.
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This means that one mole of HCl will react completely with exactly one mole of NaOH.
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However, in this experiment, we don’t have a full mole of NaOH.
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Instead, we have 0.0025 moles.
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But because of our one-to-one ratio, we know that to react with 0.0025 moles of our base, we need exactly 0.0025 moles of our acid.
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So we can simply copy the number of moles of our NaOH into the box for the number of moles of our acid.
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Now that we have 𝑛 and 𝑣 for our acid, we can rearrange our key equation and work out the concentration.
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The concentration is equal to 𝑛 divided by 𝑣.
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We can double check this by looking at the units, where the units for 𝑛 are moles, volume is liters.
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So moles divided by liters gives us moles per liter, which are the right units for concentration.
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So doing the number of moles divided by volume gives us a concentration of 0.05 molar.
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Now let’s look at an example where we need another key equation.
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28.5 grams of KOH was dissolved to make 0.2 liters of solution.
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0.067 liters of this KOH solution completely reacted with 0.12 liters of nitric acid of unknown concentration.
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What is the concentration of the acid?
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Again, let’s start by writing out a balanced equation.
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We’re told that our reactants are KOH and nitric acid with the formula for nitric acid being HNO3.
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Again, we have an acid–base reaction.
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So we’re going to produce a salt and water.
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In this case, we have KNO3 and H2O.
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And this equation is balanced.
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Let’s again try drawing out a table to help us.
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This time, however, we’ll need to be careful to pick out the right data.
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For example, we’re given two different volumes of KOH.
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So which one is the one that we need?
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Let’s take this first sentence in isolation.
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We’re told that we have 28.5 grams of KOH to dissolve into a solution of 0.2 liters.
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This sounds like it’s leading us towards a concentration of the KOH.
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And if we read the rest of the question, we’re not actually told the concentration of the KOH.
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And that’s important.
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So let’s remember another key equation.
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Here, we have 𝑛 equals 𝑚 over capital 𝑀, where 𝑛 is the number of moles, lowercase 𝑚 is mass in grams, and capital 𝑀 is molar mass in grams per mole.
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Let’s use this to work out the concentration of our KOH solution.
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The question tells us that we use 28.5 grams of KOH.
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So we can substitute this in for lowercase 𝑚.
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The molar mass we can get from our periodic table.
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We simply add together the masses of K, O, and H.
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This means that the number of moles is 28.5 grams divided by 56.105 grams per mole, which gives us 0.50798 moles.
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Now we need to return to our first key equation to work out our concentration.
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By rearranging our equation, we get concentration equals number of moles divided by volume, which works out as 0.50798 moles from where we just calculated it and 0.2 liters given in the question, leaving us with a concentration of 2.5399 molar, remembering that molar is the same as moles per liter.
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So let’s add this to our table.
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Now, we need the amount or volume of KOH solution used in the titration.
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The question tells us that this is 0.067 liters.
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It then tells us that we reacted this with 0.12 liters of acid.
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So we can add that to our table as well.
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And the question is asking us for the concentration of nitric acid.
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So now, let’s fill in as many boxes as we can until we can work out the concentration.
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Let’s start by working out the number of moles of KOH.
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We do this by multiplying the concentration by the volume, which gives us 0.17017 moles.
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Now, let’s go to our molar ratio.
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The coefficients for both KOH and HNO3 are one, so it’s a one-to-one molar ratio.
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This means that if we have 0.17017 moles of KOH, we need exactly the same number of moles of acid.
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Now that we have the number of moles, we can work out the concentration.
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We do 𝑛 divided by 𝑣, giving us a concentration of 1.418 molar.
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Now that we have a rough idea of how to do these calculations, let’s have a quick look at unit conversions.
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In our key equation of 𝑛 equals 𝑐𝑣, we’ve mainly been using the units of moles, liters, and moles per liter.
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Likewise, in our other key equation of 𝑛 equals 𝑚 over capital 𝑀, we’ve been using the units of moles, grams, and grams per mole.
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But what if the question doesn’t give our data in these specific units?
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Remember that if you see decimeters cubed, these are the same units as liters.
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And centimeters cubed are the same as milliliters.
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We’re often given values in milliliters and need to convert to liters.
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So let’s look at how to do that.
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Since there are 1000 milliliters in every liter, we can convert milliliters to liters by multiplying by one liter per 1000 milliliters.
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Remember that milli- means thousandth.
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So if we were to convert 200 milliliters into liters, we would multiply by one liter per 1000 milliliters, giving us 0.2 liters.
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Notice that by using this method, the milliliters cancel, and we’re left with liters.
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The same conversion technique works for converting milligrams into grams or millimolar into molar.
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So let’s try combining unit conversions with carrying out titration calculations.
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A 30-millimeter solution of nitric acid was titrated against a 0.1-molar solution of potassium hydroxide.
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The addition of 26.6 millimeters of potassium hydroxide was found to neutralize the nitric acid.
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What is the concentration of the nitric acid?
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Give your answer to two decimal places.
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This question is asking us to perform a titration calculation.
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This means we’re going to need the equation 𝑛 equals 𝑐𝑣, where 𝑛 is the amount in moles, 𝑐 is the concentration in moles per liter, and 𝑣 is volume in liters.
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Let’s start by working out exactly what reaction is occurring here.
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We’re told that we’re reacting nitric acid with potassium hydroxide.
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So this is an acid–base titration.
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Let’s finish writing out our reaction equation.
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When writing out our equation, we should check that it’s balanced.
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And luckily, it is.
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Now, let’s extract the important information from the question and put it into a table.
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We’re told that we need 30 milliliters of nitric acid and 26.6 milliliters of potassium hydroxide.
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So these are our values for 𝑣.
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However, currently, they’re not in the units that we want.
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To convert milliliters to liters, we multiply by one liter per 1000 milliliters.
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By converting these values for 𝑣, we get 0.03 liters of acid and 0.0266 liters of potassium hydroxide.
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Next, we’re given the concentration of the potassium hydroxide as 0.1 molar.
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Remember that molar means moles per liter.
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And the question is asking us to work out the concentration of the nitric acid.
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So let’s fill in the boxes of our table until we can work this out.
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First, we can work out the number of moles of potassium hydroxide.
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We do this by multiplying the concentration and the volume, giving us 0.00266 moles.
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But how do we know how many moles of acid we need?
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To work this out, we need to look at the molar ratio.
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We can see from our balanced chemical equation that both the nitric acid and the potassium hydroxide have a coefficient of one.
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This means we have a one-to-one molar ratio.
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So to neutralize one mole of acid, we need exactly one mole of potassium hydroxide.
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But we have 0.00266 moles of potassium hydroxide instead.
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So this means that we need exactly the same number of moles of acid.
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Now that we have 𝑛 and 𝑣 for our acid, we can work out the concentration.
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We do the number of moles divided by the volume in liters, giving us 0.0887 molar.
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However, we’re asked for two decimal places.
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So our answer is 0.09 molar or moles per liter.
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So let’s look at the key points.
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Our key equations for titration calculations are 𝑛 equals 𝑐𝑣 and 𝑛 equals 𝑚 over capital 𝑀.
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In these calculations, we use the volume of a known concentration to calculate an unknown concentration.
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The trick to doing these calculations is to use a balanced chemical equation and to carefully extract the key information, remembering to be careful with your units.