Volumetric Titrations - Online Tutor, Practice Problems & Exam Prep

Titrations are essential in stoichiometric reactions, particularly in analytical chemistry. Gravimetric analysis helps determine the amount of product isolated in a reaction, utilizing stoichiometry for calculations. A stoichiometric chart, based on a balanced chemical equation, allows for mole-to-mole comparisons using coefficients. Given information may include mass or substance units, and conversions may involve molecular weight or Avogadro's number. Understanding this framework is crucial for performing accurate titrations and stoichiometric calculations.

Volumetric Titration

Titrations are an important topic that we're going to discuss when it comes to stoichiometric reactions, but we'll talk about it consistently throughout many of the chapters as we delve deeper and deeper into analytical chemistry. Here, we're going to say that connected to the idea of titrations, we have other terms that you should recall. Here we're going to say that when we're dealing with gravimetric analysis, this is just basically helping us to determine the amount of product that we can isolate within a chemical reaction. We're going to say the calculations that we use to determine that isolated amount is termed stoichiometry. With stoichiometry, we have to employ the use of the stoichiometric chart.

Remember, with the stoometric chart, we'll be given a balanced chemical reaction and from that balanced chemical reaction, we'll be given amounts for either some elements and compounds, and asked to find the unknown quantity of some other element or compound within that balanced equation. Now with the stoichiometric chart, we have our given information which typically is given to us in some units of mass or substance. They may give us given amounts in grams, possibly moles. Remember, when we're going from grams to moles, we use the molecular weight of the compound itself. They may give us the initial amount in atoms, ions, molecules or formula units.

In this case, to go to moles, we'd use Avogadro's number which is 6.022×1023. Remember when we use the term atoms, atoms are basically for neutral single elements. So, if we're talking about oxygen atom, we're just talking about O. We use the term for ions when we have a single element with a charge. Then, the term molecules is used for covalent compounds, which are just nonmetals connected together.

And formula units are used for ionic compounds where we have a positive ion connected to a negative ion. Now, using the stoichiometric chart, we'll be given information. Usually one of these values, and it's our job to go to the unknown amount of our other compound or element within the balanced equation. Now, when we're looking for the unknown, they may ask us to find the unknown in moles, atoms, ions, molecules, formula units, or grams itself. Remember, as we're going from an area where we know stuff because it's given to us, to an area where we know nothing at all, we have to make basically a leap from one side to the other side.

That's why we call this the jump as we go from the side where we are given information to the side where we have our unknown. Remember, when we make this jump, we have to do a mole-to-mole comparison. That's why a balanced equation is needed with stoichiometry because when we do this mole-to-mole comparison, we have to use the coefficients within the balanced equation. Knowing this basic setup for the stoichiometric chart will be essential for any type of titration, stoichiometric calculations that we're going to employ. Now that we've looked at that chart, come back, click on the following video and take a look at how I answer the following example question that's left on the bottom by using the stoichiometric chart and what we remember dealing with stoichiometry.

Volumetric Titration

So let's take a look at the example given below. Here it says that iron(III) can be oxidized by an acidic potassium dichromate solution according to the net ionic equation below. We're asked how many microliters of a 0.250 molar iron(II) chloride solution are needed to completely react with 9.12 grams of a compound containing 41.5 percent potassium dichromate. Alright. Remember, with any type of board problem, even stoichiometric ones, we always start out with what exactly they are asking me to find. We're being asked to find microliters of my iron(II) chloride solution. Next, we write down all the given information that we have. From the information that we can see, we have 0.250 molar iron(II) chloride. What is that really saying? Remember, molarity is moles over liters. So that's really saying that we have 0.250 moles of iron(II) chloride over 1 liter. We have 9.12 grams of a compound. And it's saying that of that 9.12 grams of that compound, 41.5 weight percent of it is potassiu

Titrations Calculations 1

We continue our discussion of titrations with the following example question. Here it says magnesium reacts with hydrochloric acid according to the reaction below. We are asked how many grams of 5.310% by weight of aqueous magnesium are required to provide a 25% excess to react with 75 mL of 0.0550 molar hydrochloric acid. In this question, we are looking to find the grams of the aqueous magnesium solution.

From the information provided, we need to isolate that variable. They indicate 5.310 percent, meaning we have 5.310 grams of magnesium for every 100 grams of solution. We have 75 mL of 0.0550 molar hydrochloric acid.

Recall that 'of' means to multiply. We'll convert mL to liters, knowing that 1 liter equals 1000 milliliters. The molarity translates to 0.0550 moles of hydrochloric acid over 1 liter of solution. We derive that this gives us 0.004125 moles of hydrochloric acid. This quantity represents our 100% amount of HCl. The question requires a 25% excess, thus we multiply these moles by 1.25, resulting in 0.005156 moles of HCl for a 25% excess.

Now, we need to convert this into grams of solution. Using stoichiometry, we note that for every 1 mole of magnesium, there are 2 moles of hydrochloric acid. Applying the periodic table, 1 mole of magnesium weighs 24.3050 grams. Considering the weight percentage, we calculate 5.310100 grams of magnesium in the solution. This leads us to determine that the total solution weighs about 1.18 grams, rounded to 1.2 grams based on significant figures.

Thus, we conclude that 1.2 grams of the solution are necessary for the titration with a 25% excess. This is a typical type of titration problem. It's essential to first identify what you are being asked to find, then utilize all the given information effectively.

Titrations Calculations 1

So continuing with our discussion of volumetric titrations, we have example 2. Here it states the density of 2.20 molar solution of methanol is 0.976 grams per milliliter. Here it asks us what is the molality of the solution? The molar mass of methanol is given as 31.034 grams per mole. Now remember that molality is lowercase m.

Now molality equals our moles of solute divided by kilograms of solvent. This is going to require a bit of work on our part in order to isolate moles of solute and kilograms of solvent. From the information given, we see that we have a 2.20 molar solution. Remember, molarity itself equals moles of solute over liters of solution. So this 2.20 molar is equivalent to 2.20 moles of solute, which would be methanol, divided by 1 liter of solution.

Also, we're given the density of our solution. The density of our solution, denoted as d, is 0.976 grams of solution per 1 milliliter of solution. The molar mass is given as well, but we could calculate that on our own, so you don't even need to consider that given information at this point. Without even doing any math, we can see that we can isolate the moles of solute because for molarity, we have the moles of our solute, which is methanol. So we already have that part figured out.

The hard part comes from identifying your kilograms of solvent. To do that, realize that we have the volume of our solution, which is 1 liter, and we have the density of our solution. We're going to take those 1 liters of solution and say that they are equivalent to 1,000 milliliters of solution. Then I'm going to bring in the density. So, 1 milliliter of solution here on the bottom, and then we have 0.976 grams of solution on top. When we multiply those together, that's going to give me 976 grams of solution. Remember, a solution is comprised of two parts: it's made up of a solute and a solvent together. That means if I can isolate the solute, what I'll have left is grams of solvent. Remember, our solute is methanol.

We have the moles of methanol from the very beginning. I'm going to convert those into grams using the molar mass, so it's 1 mole of methanol for every 31.034 grams of methanol. When we do that, that's going to give me a mass of 68.2748 grams of methanol. Subtracting it from the 976 will give us our grams of solvent, which turns out to be 907.7252 grams of solvent. Now that we have those grams of solvent, just change them into kilograms. So we have 907.725 grams of solvent and we're going to say here for every 1 kilogram, there are 1,000 grams. That's 0.907725 kilograms of my solvent. Divide the moles of solute by the kilograms of solvent to find the molality:

2.20 ÷ 0.907725 = 2.42

So that'll represent the molality of our solution. Remember, we can isolate molality of a solution if we have the density of the solution and the molarity of the solution.

Use these approaches and you'll be able to answer any questions similar to this one.

Titrations Calculations 2

In this video, we take a look at a back titration. Now, in a back titration, we have 2 compounds reacting with one another and at the end, we're going to have an excess remaining of one of the reagents. With that excess, we then react it with a second compound. This can help us determine the concentration of the initial analyte or it can help us to figure out the volume of the second compound being used. The variety of questions from a back titration can be endless depending on how you write it.

In this first example, we will take a look at a typical type of back titration question. Here it says 0.4317 grams sample of calcium carbonate is added to a flask that also contained 12.50 ml of 1.530 molar hydrobromic acid. Now here we have our balanced equation for hydrobromic acid. It says additional water is then added to create a 250 ml of solution A. Next, 20 ml aliquot of solution A is taken and titrated with 0.0980 molar sodium hydroxide.

We're asked at the end how many milliliters of sodium hydroxide were used. Alright, so in this question, we need to figure out what our ml of sodium hydroxide are. And, what we need to realize here is that we have potassium carbonate reacting with hydrobromic acid. And we're going to say here if we were the first step we're going to do is just set up a regular stoichiometric example. We're going to say here that we have 0.4317 grams of calcium carbonate.

We're going to say here for every 1 mole of calcium carbonate, how many grams of calcium carbonate do we have in that? Well, we say that calcium we have 1 calcium, 1 carbon, and 3 oxygens. Based on the periodic table, their masses are 40.078 grams, 12.0107 grams, and 15.99994 grams. Multiply across, add up the totals, gives us 100.087 grams of calcium carbonate. At this point, we're now going to see what the stoichiometric relationship is between calcium carbonate and hydrobromic acid.

So here we're going to say according to my balanced equation, for every one mole of calcium carbonate, there are 2 moles of hydrobromic acid. All we're doing at this point is just figuring out stoichiometrically how many moles of hydrobromic acid would be needed to completely react with this amount of calcium carbonate. That's all we're doing. So for that many grams of calcium carbonate, we need exactly 0.008627 moles of hydrobromic acid. What we have to do now is we have to figure out from this volume and this molarity how many moles of HBr do we actually have present.

So we're going to change our ml's into liters. Remember, your molarity is moles over liters. So, that comes out to 0.019125 moles of hydrobromic acid present. So you can clearly see that the amount of HBr needed is smaller than the amount of HBr that we have present. That really asks us now how much of the HBr was left unused.

We have this initial amount of hydrobromic acid minus what we need This is how much hydrobromic acid we have left from the first equation. This is how many moles of HBr we have left and we're told that water is added to these moles of HBr left until we have a volume of 250 ml's. That's saying that we have 0.01048 moles of HBr and it's connected to 0.250 liters. All I did here is I converted ml's into liters. Next, they say that we take 20 ml aliquot of that amount.

Okay. So all I'm doing is I'm taking 20 ml's of that HBr. What does that mean? 20 ml's when I change it to liters is 0.020 liters. That portion is just telling me to multiply those 2 together to figure out how many moles of HBr I took out from just that 20 ml aliquot portion.

And remember, we're trying to figure out how many ml's of NaOH we have. We have the molarity of NaOH. So what we're going to do here is we're going to say, according to this second equation, for every one mole of NaOH, I have 1 mole of HBr. So for every one mole of HBr, we have 1 mole of NaOH. Then all I have to do at this point is I'm going to use the molarity now that's right here, which is 0.0980 moles of NaOH for every 1 liter.

And for every 1 liter, we have a 1,000 milliliters. So at the end, that gives me 8.57 ml's. And here we're going to do 3 significant figures because this number here has 3 significant figures compared to the others which have 4. So this is 8.57 ml of NaOH would have been needed in order to completely neutralize any excess HBr left.

This is a little bit different from what we're accustomed to seeing with titration questions. Remember in a back titration question, we utilize 2 different equations. We'll have some excess of an analyte left so we'll have to use a second compound in order to neutralize the remainder of that analyte. Doing that can help us to determine the molarity of the second compound, its volume, even its molar mass. In this example, we use that excess HBr and compared it to the NaOH in order to figure out the volume of the NaOH that was present.

Titrations Calculations 2

Here is an example to illustrate: the amount of iron can be determined by using dichromate to oxidize Iron 2 to Iron 3. The equation for this process is shown below:

6 Fe2++Cr2O72-+14H⁺→6 Fe3++2Cr³⁺+7HOl

This example asks us to determine the percentage by mass of iron in a 0.3500 grams ore sample if the complete titration of Iron 2 requires 20.7 milliliters of a 0.0185 molar dichromate solution. To find the percent by mass of iron, the calculation would be as follows:

Percent by mass of iron = (grams of iron, in terms of Iron 2 ion, divided by the grams of the ore sample) × 100.

We are given 0.3500 grams of the ore sample. We use solution stoichiometry to determine the grams of Iron 2 ion. First, convert milliliters to liters (ml/1000), giving 0.0207 liters. Multiply this by 0.0185 molarity of dichromate, which gives:

0.0207×0.0185= 3.8295×10-4 moles of dichromate ion

Using stoichiometry with the coefficients from the balanced equation (1 mole of dichromate ion yields 6 moles of Iron 2 ion), we calculate the moles of Iron 2 ion:

6× 3.8295×10-4 moles of Iron 2 ion

Now, using the molar mass of iron (55.85 grams per mole for the ion), the grams of Iron 2 ion are:

3.8295×10-4×55.85=0.1283 grams of Iron 2 ion

Finally, calculating the percent by mass of iron in the ore sample gives:

0.1283 0.3500 ×100=36.7%

This 36.7 percent is the percent by mass of iron within this ore sample, which will be our final answer.

Titrations Calculations

For this example question, it asks, "What is the molar mass of a 0.750 gram sample of a diprotic binary acid if it requires 50 amounts of 0.440 molar calcium hydroxide to completely neutralize it?" Alright. So, first of all, we're dealing with an acid and a base mixing. Consequently, we have a titration; therefore, we need our balanced equation. Here they mention "diprotic," which means it has 2 acidic hydrogens, so H₂.

Binary just means it doesn't have oxygen present, so here our placeholder, since we don't know what it is, would be A. So, H₂A is the generic formula for a diprotic binary acid. Then, if we add calcium hydroxide, it would result in the following reaction:

H 2 A + Ca(OH) 2 → 2 H O + CaA

Now our equation is balanced. All these other coefficients are just 1. Now they're asking for the molar mass of this diprotic binary acid. Remember, molar mass itself is grams per mole. You already know you possess 0.750 grams of this acid.

To figure out the moles, we'll utilize stoichiometry. Recall that molarity equals moles over liters and multiply both to obtain moles:

Moles equals liters times molarity.

Here we do the calculations:

50 mL = 50 × 10 - 3 liters × 0.440 moles/liter 50 \, \text{mL} = 50 \times 10^{-3} \, \text{liters} \times 0.440 \, \text{moles/liter}

This cancels out with the liters unit, and when plugged in, it gives the moles of the diprotic acid:

0.022 moles 0.022 \, \text{moles}

Take the 0.750 grams started with and divide by 0.022 moles to calculate the molar mass:

34.09 g/mol 34.09 \, \text{g/mol}

This particular diprotic binary acid has a molar mass of 34.09 g/mol. This result is our final answer.

A 1.000 g sample of Na₂CO₃ (MW: 105.99 g/mol) is dissolved in enough water to make 200.0 mL of solution. A 25.00 mL aliquot required 32.18 mL of HCl to completely neutralize it. What is the molar concentration of HCl? 

Na₂CO₃ (aq)  + 2 HCl (aq)  →  2 KCl (aq)  +  H₂O(l)  +  CO₂  (g)

• A. 0.5864 M
• B. 0.07330 M
• C. 0.01832 M
• D. 0.002359 M