Volumetric Analysis: Videos & Practice Problems

Concentration of solutions can be expressed through weight percent, volume percent, and weight volume percent. Weight percent is calculated as the mass of solute divided by the mass of the solution, multiplied by 100. Volume percent uses the volume of solute over the volume of solution, also multiplied by 100. Weight volume percent combines both, using the mass of solute in grams over the volume of solution in milliliters, multiplied by 100. Understanding these concepts allows for interconversion with molarity and molality, enhancing comprehension of solution chemistry.

Volumetric Analysis Concept 1

In addition to molarity and molality, the concentration of solutions can also be expressed through weight percent, volume percent, and weight volume percent. Each of these methods provides a different perspective on how solutes are measured within a solution.

Weight percent, also known as mass percent, is calculated using the formula:

\[\text{Weight Percent} = \left( \frac{\text{Mass of Solute}}{\text{Mass of Solution}} \right) \times 100\]

This formula indicates the mass of the solute relative to the total mass of the solution, expressed as a percentage.

Volume percent focuses on the volume of the solute in relation to the total volume of the solution. The formula for volume percent is:

\[\text{Volume Percent} = \left( \frac{\text{Volume of Solute}}{\text{Volume of Solution}} \right) \times 100\]

Here, both the solute and solution volumes are typically measured in milliliters.

Weight volume percent combines aspects of both weight and volume, calculated as follows:

\[\text{Weight Volume Percent} = \left( \frac{\text{Mass of Solute (g)}}{\text{Volume of Solution (mL)}} \right) \times 100\]

This measure indicates how many grams of solute are present in a specific volume of solution.

Understanding these three additional methods of expressing concentration allows for greater flexibility in solving problems related to solutions. These concepts can be interconverted with molarity and molality through various manipulations, enhancing the ability to analyze and work with different types of concentration measurements effectively.

Volumetric Analysis Example 1

To determine the weight percent (or mass percent) of potassium sulfate in a solution, we start with the formula:

Weight percent = \(\frac{\text{mass of solute}}{\text{mass of solution}} \times 100\)

In this case, the mass of potassium sulfate (the solute) is given as 0.30 grams. To find the mass of the solution, we need to use the volume of the solution and its density. The volume of the solution is 5.12 liters, and the density is 1.30 grams per milliliter.

First, we convert the volume from liters to milliliters:

5.12 liters = 5.12 × 1000 = 5120 milliliters

Next, we calculate the mass of the solution using the density:

Mass of solution = Volume × Density = 5120 mL × 1.30 g/mL = 6656 grams

Now, we can substitute the values into the weight percent formula:

Weight percent = \(\frac{0.30 \text{ g}}{6656 \text{ g}} \times 100\)

Calculating this gives:

Weight percent = \(\frac{0.30}{6656} \times 100 \approx 0.0045\%\)

To express this in scientific notation, we write:

Weight percent ≈ \(3.46 \times 10^{-3}\%\)

Considering significant figures, the final answer is:

Weight percent of potassium sulfate = \(3.46 \times 10^{-3}\%\)

Parts per million of Lead

Understanding lead levels in blood is crucial for health assessments, particularly when levels exceed 0.80 parts per million (ppm), which equates to 0.80 grams of lead in 1,000,000 grams of blood. To determine the grams of lead in 550 milliliters of blood with a lead concentration of 0.583 ppm, we can follow a systematic approach to convert and calculate the necessary values.

First, we recognize that 0.583 ppm indicates that there are 0.583 grams of lead in 1,000,000 grams of blood. Given that the density of blood is 1,060.0 kilograms per cubic meter, we need to convert our volume of blood from milliliters to grams. Since 1 milliliter is equivalent to 1 cubic centimeter, we can directly relate milliliters to grams using the density.

To convert the volume of blood, we start with 550 milliliters. We convert this volume to grams using the density:

Density of blood = 1,060.0 kg/m³ = 1,060.0 g/L (since 1 m³ = 1,000 L)

Thus, the mass of 550 mL of blood can be calculated as follows:

Mass of blood = Volume × Density = 550 mL × 1,060 g/L = 583 g

Next, we can calculate the grams of lead present in this mass of blood using the conversion factor derived from the ppm value:

Grams of lead = (0.583 g lead / 1,000,000 g blood) × 583 g blood

Calculating this gives:

Grams of lead = 0.583 × 583 / 1,000,000 = 0.00033989 g

To express this result in scientific notation and with the appropriate number of significant figures, we note that the values used in the calculation have varying significant figures. The final answer should reflect the least number of significant figures from the values used, which in this case is three. Therefore, we round the result to:

Final grams of lead = 3.40 × 10^-4 grams

This problem illustrates the importance of unit conversion and dimensional analysis in chemistry, allowing us to systematically derive the desired quantity from given data. By organizing the information and carefully canceling units, we can arrive at a clear and accurate answer.

Percent Composition

To calculate the molar concentration of sulfate ions in an 8.13% aluminum sulfate solution with a density of 1.235 grams per milliliter, we start by understanding the composition of the solution. An 8.13% solution means there are 8.13 grams of aluminum sulfate in every 100 grams of solution. Given the density, we can convert the mass of the solution into volume, which is essential for determining molarity.

First, we convert the grams of aluminum sulfate to moles using its molar mass. The molar mass of aluminum sulfate, \(\text{Al}_2(\text{SO}_4)_3\), is approximately 342.17 grams per mole. The conversion is as follows:

Number of moles of aluminum sulfate = \(\frac{8.13 \text{ g}}{342.17 \text{ g/mol}} \approx 0.0237 \text{ moles}\)

Next, we recognize that each mole of aluminum sulfate produces three moles of sulfate ions. Therefore, the moles of sulfate ions can be calculated as:

Moles of sulfate ions = \(0.0237 \text{ moles Al}_2(\text{SO}_4)_3 \times 3 \approx 0.0711 \text{ moles SO}_4^{2-}\)

Now, we need to find the volume of the solution in liters. Using the density of the solution, we can convert the mass of the solution to volume:

Volume of solution = \(\frac{100 \text{ g}}{1.235 \text{ g/mL}} \approx 81.0 \text{ mL}\)

Converting milliliters to liters gives us:

Volume of solution in liters = \(\frac{81.0 \text{ mL}}{1000} = 0.0810 \text{ L}\)

Finally, we can calculate the molarity of sulfate ions by dividing the moles of sulfate ions by the volume of the solution in liters:

Molarity of sulfate ions = \(\frac{0.0711 \text{ moles SO}_4^{2-}}{0.0810 \text{ L}} \approx 0.878 \text{ M}\)

Rounding to three significant figures, the final molarity of sulfate ions in the solution is approximately 0.878 M.

Percent Composition

To determine the volume of a sodium acetate solution needed for a reaction requiring a specific amount of the compound, we start with the given data: a 33.8% sodium acetate solution with a density of 1.10 grams per milliliter, and a requirement of 68.8 grams of sodium acetate. However, since we want to use a 50% excess of sodium acetate, we first need to calculate the total amount required.

To find the total sodium acetate needed, we convert the 50% excess into a decimal by adding 1.50 (which represents 150% of the theoretical yield). Thus, we calculate:

\[\text{Total Sodium Acetate} = 1.50 \times 68.8 \, \text{grams} = 103.2 \, \text{grams}\]

Next, we need to convert grams of sodium acetate into milliliters of solution. We know that in a 33.8% solution, there are 33.8 grams of sodium acetate per 100 grams of solution. Therefore, we can set up a conversion factor:

\[\frac{103.2 \, \text{grams NaAc}}{33.8 \, \text{grams NaAc}} \times \frac{100 \, \text{grams solution}}{1} \]

Now, we need to convert grams of solution to milliliters using the density of the solution (1.10 grams/mL):

\[\frac{1 \, \text{mL}}{1.10 \, \text{grams solution}} \]

Combining these conversions, we can isolate the milliliters of solution:

\[\text{Volume (mL)} = 103.2 \, \text{grams NaAc} \times \frac{100 \, \text{grams solution}}{33.8 \, \text{grams NaAc}} \times \frac{1 \, \text{mL}}{1.10 \, \text{grams solution}} \]

Calculating this gives:

\[\text{Volume (mL)} \approx 277.569 \, \text{mL}\]

Rounding to three significant figures, the final volume of the sodium acetate solution required is approximately 278 mL. This problem illustrates the importance of understanding how to calculate excess amounts in chemical reactions and how to convert between different units effectively.

Determine the number of mmoles of sulfuric acid that would be found within a 120 g sample that is 79.9% sulfuric acid.

• A. 0.960 mmol 
• B. 980 mmol
• C. 0.33 mmol
• D. 1.7 mmol

Answer: (B)

Dilution

A dilution is a process that involves adding water to a concentrated solution, effectively reducing its concentration. In practice, this is often done by placing a specific volume of the concentrated solution into a volumetric flask and then adding water up to a predetermined mark. The relationship between the concentrations and volumes before and after dilution is described by the dilution equation:

\( m_1 V_1 = m_2 V_2 \)

In this equation, \( m_1 \) represents the molarity of the concentrated solution before dilution, while \( V_1 \) is the initial volume of that concentrated solution. Conversely, \( m_2 \) denotes the molarity of the diluted solution, and \( V_2 \) is the final volume after water has been added. It is important to note that \( m_1 \) is typically greater than \( m_2 \), and \( V_2 \) is greater than \( V_1 \). The final volume \( V_2 \) can be expressed as the sum of the initial volume \( V_1 \) and the volume of water added.

Understanding this dilution process and the associated formula is crucial for solving related problems. When you encounter examples, remember that the addition of water signifies a dilution, allowing you to apply the equation effectively. If you find yourself struggling with a specific example, seeking additional guidance can be beneficial.

Dilutions

To determine how many grams of a 53.1 weight percent sodium chloride (NaCl) solution are needed to prepare 2.50 liters of a 0.15 M NaCl solution, we start by understanding the relationship between molarity, volume, and the mass of solute required.

Molarity (M) is defined as the number of moles of solute per liter of solution. In this case, a 0.15 M NaCl solution means there are 0.15 moles of NaCl in every liter of solution. Therefore, for 2.50 liters, the total moles of NaCl needed can be calculated as:

\[\text{Moles of NaCl} = \text{Molarity} \times \text{Volume} = 0.15 \, \text{mol/L} \times 2.50 \, \text{L} = 0.375 \, \text{moles}\]

Next, we convert moles of NaCl to grams using the molar mass of NaCl, which is approximately 58.443 g/mol:

\[\text{Grams of NaCl} = \text{Moles} \times \text{Molar Mass} = 0.375 \, \text{moles} \times 58.443 \, \text{g/g} \approx 21.9 \, \text{grams}\]

Now, since the sodium chloride solution is 53.1 weight percent, this means that in every 100 grams of solution, there are 53.1 grams of NaCl. To find the total mass of the solution needed to obtain 21.9 grams of NaCl, we set up a proportion:

\[\frac{53.1 \, \text{g NaCl}}{100 \, \text{g solution}} = \frac{21.9 \, \text{g NaCl}}{x \, \text{g solution}}\]

Cross-multiplying gives:

\[53.1 \, \text{g NaCl} \cdot x = 21.9 \, \text{g NaCl} \cdot 100 \, \text{g solution}\]

Solving for \(x\):

\[x = \frac{21.9 \, \text{g NaCl} \cdot 100 \, \text{g solution}}{53.1 \, \text{g NaCl}} \approx 41.2 \, \text{g solution}\]

Rounding to three significant figures, we find that approximately 41 grams of the sodium chloride solution are required to achieve the desired concentration. This calculation illustrates the importance of understanding the relationships between molarity, mass, and weight percent in solution preparation.

If 920 mL of water is added to 78.0 mL of a 1.28 M HBrO₄ solution what is the resulting molarity? 

• A. 0.110 M
• B. 15.1 M
• C. 0.100 M
• D. 0.119 M

Answer: (C)

Volumetric Analysis Example 6

To determine the final concentration of sodium hydroxide (NaOH) after a series of dilutions, we start with the preparation of a stock solution. A student dissolved 25 grams of NaOH in enough water to make 150 mL of solution. Given that the molecular weight of sodium hydroxide is 40 grams per mole, we can calculate the number of moles in the stock solution:

Number of moles of NaOH = \(\frac{25 \text{ grams}}{40 \text{ grams/mole}} = 0.625 \text{ moles}\).

Next, we find the initial molarity (M₁) of the stock solution. Molarity is defined as moles of solute per liter of solution:

M₁ = \(\frac{0.625 \text{ moles}}{0.150 \text{ liters}} = 4.167 \text{ M}\).

Now, the student took 20 mL of this stock solution and diluted it to 250 mL. To find the new molarity (M₂), we use the dilution formula:

M₁V₁ = M₂V₂,

where V₁ is the volume of the stock solution (20 mL) and V₂ is the final volume after dilution (250 mL). Rearranging gives us:

M₂ = \(\frac{M₁V₁}{V₂

Next, the student took 75 mL of this new solution and diluted it to 500 mL. Again, we apply the dilution formula:

M₂V₂ = M₃V₃,

where V₂ is 75 mL and V₃ is 500 mL. Rearranging gives us:

M₃ = \(\frac{M₂V₂}{V₃

Considering significant figures, the final concentration of sodium hydroxide after these dilutions is reported as 0.05 M.

Dilutions

To prepare a 0.425 M sodium hydroxide (NaOH) solution, we need to determine how many milliliters of a 63.7 weight percent NaOH solution are required to dilute to a final volume of 850 mL. The density of the sodium hydroxide solution is given as 0.915 grams per milliliter.

First, we convert the desired volume of the solution from milliliters to liters, since molarity (M) is defined as moles of solute per liter of solution. Thus, 850 mL is equivalent to 0.850 L. Using the molarity formula:

Molarity (M) = Moles of solute / Volume of solution (L)

We can calculate the moles of NaOH needed:

Moles of NaOH = Molarity × Volume = 0.425 mol/L × 0.850 L = 0.36125 moles

Next, we need to convert moles of NaOH to grams. The molar mass of NaOH is calculated by summing the atomic masses of sodium (Na), oxygen (O), and hydrogen (H):

Molar mass of NaOH = 22.98977 g (Na) + 15.9994 g (O) + 1.00794 g (H) = 39.9971 g/mol

Now, we can find the mass of NaOH required:

Mass of NaOH = Moles × Molar mass = 0.36125 moles × 39.9971 g/mol = 14.445 g

Given that the solution is 63.7 weight percent NaOH, this means there are 63.7 grams of NaOH in every 100 grams of solution. To find the total mass of the solution needed to obtain 14.445 g of NaOH, we set up the following proportion:

63.7 g NaOH / 100 g solution = 14.445 g NaOH / x g solution

Solving for x gives:

x = (14.445 g NaOH × 100 g solution) / 63.7 g NaOH = 22.66 g solution

Next, we convert the mass of the solution to volume using the density:

Density = Mass / Volume

Rearranging gives us:

Volume = Mass / Density

Substituting the values:

Volume = 22.66 g / 0.915 g/mL = 24.77 mL

Rounding to three significant figures, we find that approximately 24.8 mL of the 63.7 weight percent NaOH solution is required to be diluted to 850 mL to achieve a 0.425 M NaOH solution. This calculation illustrates the importance of carefully isolating units and applying the correct formulas to navigate through dilution problems effectively.

Here’s what students ask on this topic:

What is the difference between molarity and molality?

Molarity (M) is defined as the number of moles of solute per liter of solution. It is expressed as:

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

Molality (m) is defined as the number of moles of solute per kilogram of solvent. It is expressed as:

$m=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{kilograms} \mathrm{of} \mathrm{solvent}}$

The key difference is that molarity depends on the volume of the solution, which can change with temperature, while molality depends on the mass of the solvent, which remains constant regardless of temperature changes.

How do you calculate weight percent in a solution?

Weight percent (also known as mass percent) is calculated by dividing the mass of the solute by the mass of the solution and then multiplying by 100. The formula is:

$\mathrm{Weight} \mathrm{percent}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{mass} \mathrm{of} \mathrm{solution}} \times 100$

For example, if you have 5 grams of solute in 100 grams of solution, the weight percent would be:

$\frac{5}{100} \times 100=5\%$

What is volume percent and how is it calculated?

Volume percent is a way to express the concentration of a solution, calculated by dividing the volume of the solute by the volume of the solution and then multiplying by 100. The formula is:

$\mathrm{Volume} \mathrm{percent}=\frac{\mathrm{volume} \mathrm{of} \mathrm{solute}}{\mathrm{volume} \mathrm{of} \mathrm{solution}} \times 100$

For example, if you have 10 mL of solute in 100 mL of solution, the volume percent would be:

$\frac{10}{100} \times 100=10\%$

How do you convert between molarity and weight percent?

To convert between molarity (M) and weight percent, you need the density of the solution and the molar mass of the solute. The steps are:

1. Calculate the mass of the solute using molarity and volume of the solution.

2. Determine the total mass of the solution using its density.

3. Use the weight percent formula:

$\mathrm{Weight} \mathrm{percent}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{mass} \mathrm{of} \mathrm{solution}} \times 100$

For example, if you have a 1 M solution of NaCl (molar mass = 58.44 g/mol) and the density of the solution is 1.05 g/mL, for 1 L of solution:

Mass of NaCl = 1 mol × 58.44 g/mol = 58.44 g

Mass of solution = 1 L × 1.05 g/mL × 1000 mL/L = 1050 g

Weight percent = (58.44 g / 1050 g) × 100 ≈ 5.56%

What is weight volume percent and how is it used?

Weight volume percent (w/v%) is a concentration measure that combines the mass of the solute and the volume of the solution. It is calculated by dividing the mass of the solute (in grams) by the volume of the solution (in milliliters) and then multiplying by 100. The formula is:

$\mathrm{Weight} \mathrm{volume} \mathrm{percent}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{volume} \mathrm{of} \mathrm{solution}} \times 100$

For example, if you have 5 grams of solute in 100 mL of solution, the weight volume percent would be:

$\frac{5}{100} \times 100=5\%$

This measure is commonly used in medical and biological contexts, such as preparing solutions for intravenous fluids.