Volumetric Titrations: Videos & Practice Problems

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 play a crucial role in stoichiometric reactions and are a recurring theme in analytical chemistry. Understanding the concept of gravimetric analysis is essential, as it helps determine the amount of product that can be isolated from a chemical reaction. The calculations involved in this process are referred to as stoichiometry, which utilizes a stoichiometric chart derived from a balanced chemical equation.

The stoichiometric chart serves as a tool to relate known quantities of reactants or products to unknown quantities within a balanced equation. Typically, the given information may be presented in various units such as grams or moles. To convert grams to moles, the molecular weight of the compound is used. When dealing with atoms, ions, molecules, or formula units, Avogadro's number, \(6.022 \times 10^{23}\), is applied to transition to moles. It is important to note the distinctions between these terms: "atoms" refers to neutral single elements (e.g., O for oxygen), "ions" denotes charged single elements, "molecules" pertains to covalent compounds (nonmetals), and "formula units" describes ionic compounds (combinations of positive and negative ions).

In practice, the stoichiometric chart allows us to navigate from known values to unknown quantities, which may be expressed in moles, atoms, ions, molecules, formula units, or grams. This transition is often referred to as making a "jump" from the known to the unknown. To facilitate this leap, a mole-to-mole comparison is essential, relying on the coefficients from the balanced equation. This foundational understanding of the stoichiometric chart is vital for performing titrations and other stoichiometric calculations effectively.

Volumetric Titration

In this problem, we are tasked with determining the volume in microliters of a 0.250 M iron(II) chloride solution required to completely react with 9.12 grams of a compound that contains 41.5% potassium dichromate by weight. To solve this, we begin by identifying the key information provided and the relationships between the substances involved.

First, we convert the weight percent of potassium dichromate into grams. Since 41.5% of the compound is potassium dichromate, we can calculate the mass of potassium dichromate in the 9.12 grams of the compound:

Mass of potassium dichromate = 9.12 g × (41.5 g K₂Cr₂O₇ / 100 g compound) = 3.78 g K₂Cr₂O₇

Next, we need to convert grams of potassium dichromate to moles. The molar mass of potassium dichromate (K₂Cr₂O₇) is calculated as follows:

Molar mass = 2(39.0983 g/mol) + 2(51.9961 g/mol) + 7(15.99994 g/mol) = 294.184 g/mol

Now, we can find the number of moles of potassium dichromate:

Number of moles of K₂Cr₂O₇ = 3.78 g / 294.184 g/mol = 0.0128 mol

According to the balanced chemical equation, 1 mole of potassium dichromate reacts with 6 moles of iron(II) ions. Therefore, we can calculate the moles of iron(II) required:

Moles of Fe²⁺ = 0.0128 mol K₂Cr₂O₇ × (6 mol Fe²⁺ / 1 mol K₂Cr₂O₇) = 0.0768 mol Fe²⁺

Next, we convert moles of iron(II) to moles of iron(II) chloride (FeCl₂), noting that 1 mole of Fe²⁺ corresponds to 1 mole of FeCl₂:

Moles of FeCl₂ = 0.0768 mol

Using the molarity of the iron(II) chloride solution, we can find the volume in liters:

Volume (L) = Moles of FeCl₂ / Molarity = 0.0768 mol / 0.250 mol/L = 0.3072 L

To convert this volume to microliters, we use the conversion factor where 1 L = 1,000,000 µL:

Volume (µL) = 0.3072 L × 1,000,000 µL/L = 307,200 µL

Finally, rounding to three significant figures, the required volume of the iron(II) chloride solution is:

3.07 × 10⁵ µL

This problem illustrates the importance of dimensional analysis and stoichiometric relationships in chemical calculations, allowing us to systematically convert between units and substances to arrive at the desired answer.

Titrations Calculations 1

In this example, we explore a titration involving magnesium and hydrochloric acid. The goal is to determine the grams of a 5.310% by weight aqueous magnesium solution needed to provide a 25% excess for a reaction with 75 mL of 0.0550 M hydrochloric acid.

First, we interpret the 5.310% concentration, which indicates that there are 5.310 grams of magnesium in every 100 grams of the solution. To find the moles of hydrochloric acid, we convert the volume from milliliters to liters, knowing that 1 liter equals 1000 milliliters. Thus, 75 mL converts to 0.075 L. The molarity (0.0550 M) tells us that there are 0.0550 moles of HCl per liter of solution. Therefore, the moles of HCl in 75 mL can be calculated as:

$$ \text{Moles of HCl} = 0.075 \, \text{L} \times 0.0550 \, \text{mol/L} = 0.004125 \, \text{moles} $$

Next, to account for the 25% excess, we calculate the total moles of HCl required:

$$ \text{Total moles of HCl} = 0.004125 \, \text{moles} \times 1.25 = 0.005156 \, \text{moles} $$

To find the moles of magnesium needed, we use the stoichiometric relationship from the balanced chemical equation, which states that 1 mole of magnesium reacts with 2 moles of hydrochloric acid. Thus, we can find the moles of magnesium required:

$$ \text{Moles of Mg} = \frac{0.005156 \, \text{moles HCl}}{2} = 0.002578 \, \text{moles Mg} $$

Next, we convert moles of magnesium to grams using its molar mass, which is approximately 24.3050 g/mol:

$$ \text{Grams of Mg} = 0.002578 \, \text{moles} \times 24.3050 \, \text{g/mol} = 0.0627 \, \text{grams} $$

Now, we need to relate the grams of magnesium back to the grams of the solution. Since the solution is 5.310% magnesium by weight, we set up the following ratio:

$$ \frac{5.310 \, \text{g Mg}}{100 \, \text{g solution}} $$

To find the total grams of solution required, we rearrange the equation:

$$ \text{Grams of solution} = \frac{100 \, \text{g solution}}{5.310 \, \text{g Mg}} \times 0.0627 \, \text{g Mg} = 1.18006 \, \text{grams of solution} $$

Considering significant figures, the final answer is rounded to two significant figures, resulting in:

$$ \text{Grams of solution} \approx 1.2 \, \text{grams} $$

This example illustrates the systematic approach to solving titration problems, emphasizing the importance of understanding the relationships between moles, mass, and concentration in chemical reactions.

Titrations Calculations 1

In volumetric titrations, understanding the relationship between molarity and molality is crucial. For a 2.20 molar solution of methanol with a density of 0.976 grams per milliliter, we can calculate the molality, denoted as m, which is defined as the number of moles of solute per kilogram of solvent.

First, we recognize that molarity (M) is expressed as:

M = \frac{moles \ of \ solute}{liters \ of \ solution}

Given that the solution is 2.20 M, this indicates there are 2.20 moles of methanol in 1 liter of solution. To find the total mass of the solution, we convert the volume from liters to milliliters (1 liter = 1000 milliliters) and use the density:

mass \ of \ solution = volume \times density = 1000 \ mL \times 0.976 \ \frac{g}{mL} = 976 \ g

Next, we need to isolate the mass of the solvent. The solution consists of both the solute (methanol) and the solvent (water). To find the mass of the solute, we convert moles of methanol to grams using its molar mass (31.034 g/mol):

mass \ of \ methanol = 2.20 \ moles \times 31.034 \ \frac{g}{mol} = 68.2748 \ g

Now, we can determine the mass of the solvent by subtracting the mass of the solute from the total mass of the solution:

mass \ of \ solvent = mass \ of \ solution - mass \ of \ methanol = 976 \ g - 68.2748 \ g = 907.7252 \ g

To convert grams of solvent to kilograms, we divide by 1000:

mass \ of \ solvent = \frac{907.7252 \ g}{1000} = 0.9077252 \ kg

Finally, we can calculate the molality:

molality (m) = \frac{moles \ of \ solute}{kilograms \ of \ solvent} = \frac{2.20 \ moles}{0.9077252 \ kg} \approx 2.42 \ m

This calculation illustrates how to derive molality from molarity and density, emphasizing the importance of understanding the components of a solution. By isolating the moles of solute and the mass of solvent, one can effectively determine the molality of any solution.

Titrations Calculations 2

Back titration is a technique used to determine the concentration of an analyte by reacting it with a known excess of a reagent, followed by titration of the unreacted reagent with a second solution. This method is particularly useful when direct titration is not feasible. In a typical back titration scenario, one compound reacts with another, leaving an excess of one reagent that can be titrated with a third compound.

Consider a scenario where a sample of calcium carbonate (CaCO₃) weighing 0.4317 grams is added to a flask containing 12.50 mL of 1.530 M hydrobromic acid (HBr). The balanced chemical equation indicates that one mole of calcium carbonate reacts with two moles of hydrobromic acid:

CaCO₃ + 2 HBr → CaBr₂ + H₂O + CO₂

To find out how much HBr is required to completely react with the calcium carbonate, we first calculate the molar mass of calcium carbonate:

Molar mass of CaCO₃ = 40.078 g (Ca) + 12.0107 g (C) + 3 × 15.99994 g (O) = 100.087 g/mol

Next, we determine the number of moles of calcium carbonate in the sample:

Number of moles of CaCO₃ = \(\frac{0.4317 \text{ g}}{100.087 \text{ g/mol}} = 0.00431 \text{ moles}\)

Using the stoichiometric relationship, we find that:

Required moles of HBr = 2 × 0.00431 = 0.00862 moles

Now, we calculate the moles of HBr present in the solution. First, convert the volume of HBr from mL to L:

Volume of HBr = 12.50 mL = 0.01250 L

Using the molarity formula (M = moles/volume), we find the moles of HBr:

Moles of HBr = 1.530 mol/L × 0.01250 L = 0.019125 moles

To find the unreacted HBr, we subtract the moles required from the moles present:

Unreacted moles of HBr = 0.019125 - 0.00862 = 0.010505 moles

After adding water to reach a total volume of 250 mL, we can determine the concentration of the remaining HBr:

Concentration of HBr = \(\frac{0.010505 \text{ moles}}{0.250 \text{ L}} = 0.04202 \text{ M}\)

Next, a 20 mL aliquot of this solution is taken for titration with sodium hydroxide (NaOH) of known molarity (0.0980 M). First, convert the aliquot volume to liters:

Volume of aliquot = 20 mL = 0.020 L

Now, calculate the moles of HBr in the aliquot:

Moles of HBr in aliquot = 0.04202 mol/L × 0.020 L = 0.0008404 moles

Since the reaction between HBr and NaOH is a 1:1 stoichiometry:

NaOH required = Moles of HBr = 0.0008404 moles

To find the volume of NaOH needed, we use the molarity of NaOH:

Volume of NaOH = \(\frac{0.0008404 \text{ moles}}{0.0980 \text{ mol/L}} = 0.00857 \text{ L} = 8.57 \text{ mL}\)

Thus, 8.57 mL of NaOH is required to neutralize the excess HBr in the back titration process. This method illustrates how back titration can effectively determine the volume of a reagent needed to react with an excess of another compound, providing valuable insights into the concentrations involved in the reaction.

Titrations Calculations 2

To determine the percentage by mass of iron in an ore sample, we can utilize the oxidation of iron(II) to iron(III) using dichromate. The relevant equation for this reaction is crucial for understanding the stoichiometry involved. In this case, we are given a 0.3500 grams ore sample and the complete titration of iron(II) requires 20.7 milliliters of a 0.0185 molar dichromate solution.

The formula for calculating percent by mass is:

\[\text{Percent by mass} = \left(\frac{\text{grams of iron}}{\text{grams of ore sample}}\right) \times 100\%

We already know the mass of the ore sample is 0.3500 grams. To find the grams of iron, we first need to convert the volume of dichromate solution from milliliters to liters:

\[\text{Volume in liters} = \frac{20.7 \, \text{mL}}{1000} = 0.0207 \, \text{L}\]

Next, we calculate the moles of dichromate ion using the molarity:

\[\text{Moles of dichromate} = \text{Volume (L)} \times \text{Molarity} = 0.0207 \, \text{L} \times 0.0185 \, \text{mol/L} = 3.8295 \times 10^{-4} \, \text{moles}\]

Using stoichiometry, we can find the moles of iron(II) ion. The balanced equation indicates that 1 mole of dichromate reacts with 6 moles of iron(II):

\[\text{Moles of iron(II)} = 3.8295 \times 10^{-4} \, \text{moles dichromate} \times \frac{6 \, \text{moles iron(II)}}{1 \, \text{mole dichromate}} = 2.2977 \times 10^{-3} \, \text{moles iron(II)}\]

Now, we convert moles of iron(II) to grams using the molar mass of iron, which is 55.85 grams per mole:

\[\text{Grams of iron(II)} = 2.2977 \times 10^{-3} \, \text{moles} \times 55.85 \, \text{g/mol} = 0.1283 \, \text{grams}\]

Finally, we can calculate the percent by mass of iron in the ore sample:

\[\text{Percent by mass} = \left(\frac{0.1283 \, \text{grams}}{0.3500 \, \text{grams}}\right) \times 100\% = 36.7\%\]

Thus, the percent by mass of iron in the ore sample is 36.7%.

Titrations Calculations

To determine the molar mass of a diprotic binary acid from a titration with calcium hydroxide, we start by establishing the balanced chemical equation. A diprotic acid, represented as H₂A, contains two acidic hydrogens, while a binary acid indicates the absence of oxygen. The reaction with calcium hydroxide (Ca(OH)₂) can be expressed as:

H₂A + Ca(OH)₂ → CaA + 2H₂O

In this reaction, one mole of the diprotic acid reacts with one mole of calcium hydroxide, producing one mole of calcium salt and two moles of water. Given that 50 mL of 0.440 M calcium hydroxide is used for neutralization, we first convert the volume from milliliters to liters:

50 mL = 0.050 L

Next, we calculate the moles of calcium hydroxide using the formula:

moles = volume (L) × molarity (mol/L)

Substituting the values:

moles of Ca(OH)₂ = 0.050 L × 0.440 mol/L = 0.022 moles

Since the balanced equation shows a 1:1 mole ratio between calcium hydroxide and the diprotic acid, we conclude that there are also 0.022 moles of H₂A.

To find the molar mass of the diprotic binary acid, we use the formula:

molar mass = mass (g) / moles (mol)

Given that the mass of the acid sample is 0.750 grams, we can calculate the molar mass:

molar mass = 0.750 g / 0.022 mol ≈ 34.09 g/mol

Thus, the molar mass of the diprotic binary acid is approximately 34.09 grams per mole.

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

Answer: (B)

Here’s what students ask on this topic:

What is the purpose of a titration in analytical chemistry?

The purpose of a titration in analytical chemistry is to determine the concentration of an unknown solution by reacting it with a solution of known concentration. This process involves the gradual addition of the titrant (the solution of known concentration) to the analyte (the solution of unknown concentration) until the reaction reaches its equivalence point, where the amount of titrant added is stoichiometrically equivalent to the amount of analyte present. This allows for precise and accurate quantification of the analyte's concentration.

How do you use a stoichiometric chart in titration calculations?

A stoichiometric chart is used in titration calculations to relate the quantities of reactants and products in a balanced chemical equation. First, identify the given information, such as the mass or moles of a substance. Convert this information to moles if necessary using molecular weight or Avogadro's number. Then, use the coefficients from the balanced equation to perform a mole-to-mole comparison between the reactants and products. Finally, convert the moles of the desired substance back to the required units (e.g., grams, molecules) to find the unknown quantity.

What is the role of the equivalence point in a titration?

The equivalence point in a titration is the point at which the amount of titrant added is stoichiometrically equivalent to the amount of analyte in the solution. At this point, the reaction between the titrant and the analyte is complete. The equivalence point is crucial because it allows for the accurate determination of the analyte's concentration. It is often indicated by a sudden change in a measurable property, such as pH or color, which can be detected using indicators or pH meters.

How do you convert between grams and moles in stoichiometric calculations?

To convert between grams and moles in stoichiometric calculations, use the molecular weight of the compound. The molecular weight (molar mass) is the sum of the atomic masses of all atoms in the compound, expressed in grams per mole (g/mol). The conversion is done using the following equations:

From grams to moles:

$\frac{\mathrm{mass}}{\mathrm{molecular\; weight}}$

From moles to grams:

$\mathrm{moles}\times \mathrm{molecular\; weight}$

What is Avogadro's number and how is it used in stoichiometry?

Avogadro's number, 6.022 × 10²³, is the number of atoms, ions, molecules, or formula units in one mole of a substance. It is used in stoichiometry to convert between the number of particles and moles. For example, to find the number of molecules in a given number of moles, multiply the moles by Avogadro's number. Conversely, to find the number of moles from a given number of particles, divide the number of particles by Avogadro's number. This conversion is essential for accurate stoichiometric calculations in chemical reactions.