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Moles and molar concentration

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Intros
Lessons
  1. Expanding our moles calculations
  2. Recap - moles, molar volume and concentration
  3. Calculating concentration of a solution
  4. Example - calculating concentration
  5. Titration, an introduction
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Examples
Lessons
  1. Find the number of moles and concentration of substances used in chemical reactions.
    1. Calculate the number of moles in 100 mL of 0.2M HCl (aq)_{(aq)}
    2. Calculate the concentration of a solution of 1.7 litres of water with 0.75 moles of HCl dissolved in it.
    3. Calculate the new concentration of this solution when 3.4 extra litres of water are added to the solution.
  2. Find the number of moles and use it to find the volume of substances used in chemical reactions.
    1. Calculate the number of moles in 120mL of 0.05M H2_2SO4  (aq)_{4\; (aq)} solution.
    2. When H2_2SO4_4 and CaCO3_3 are reacted, CO2_2 gas is produced. What is the volume of CO2  (g)_{2\; (g)} produced when 120 mL of 0.1M H2_2SO4  (aq)_{4\; (aq)} is reacted with an excess amount of CaCO3_3?
    3. What volume of 0.04M HCl (aq)_{(aq)} is required to react completely to neutralise 6 g of Mg(OH)2_2?
  3. Find the number of moles and use it to find the molarity of substances used in chemical reactions.
    1. In a titration reaction, 32 mL HCl (aq)_{(aq)} of unknown concentration reacts with 25 mL of 0.5M NaOH (aq)_{(aq)}. Calculate the molarity of the hydrochloric acid being analysed.
    2. In another titration reaction, it took 14mL of NaOH (aq)_{(aq)} of unknown concentration to react with 39 mL of 0.1M H3_3PO4  (aq)_{4\; (aq)}. Calculate the molarity of the NaOH solution used.
  4. Find the number of moles and use it to find the quantities of substances used in chemical reactions.
    Consider the reaction:
    2 Al  (s)+_{\;(s)} + 2 NaON  (aq)+_{\;(aq)} + 2 H2_2 O  (l)_{\;(l)} →2 NaAlO2  (aq)+_{2\;(aq)} + 3 H2  (g)_{2\;(g)}
    1. A student needs to produce 20L of H2_2 at STP using this reaction. Excess water and aluminium is available to react with 1.5M sodium hydroxide solution. How much sodium hydroxide solution will be required?
    2. When repeating the experiment, only 500 mL of 2.2M sodium hydroxide solution is available. How many litres of hydrogen gas can be produced given this limit?
Topic Notes
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Introduction

Moles and molar concentration are fundamental concepts in chemistry, essential for understanding stoichiometry and performing accurate calculations. A mole represents a specific number of particles (6.022 x 10^23), while molar concentration describes the amount of solute in a given volume of solution. The introduction video provides a clear and concise explanation of these concepts, serving as a crucial foundation for students embarking on their chemistry journey. By grasping moles and molar concentration, learners can effectively tackle more complex chemical problems and reactions. These concepts are indispensable in various chemistry calculations, including balancing chemical equations, predicting reaction outcomes, and determining product yields. Mastering moles and molar concentration is vital for success in chemistry, as they form the basis for understanding stoichiometric relationships and solution chemistry. The video's significance lies in its ability to demystify these abstract concepts, making them accessible and applicable to real-world chemical scenarios.

Understanding Moles and Solutions

In the realm of chemistry, the concept of moles plays a crucial role in understanding and quantifying substances. A mole is a fundamental unit that represents 6.022 x 10^23 particles of a substance, known as Avogadro's number. This unit allows chemists to work with manageable quantities of atoms, molecules, or ions, bridging the gap between the microscopic world of particles and the macroscopic world we observe.

The importance of moles in chemistry cannot be overstated. They serve as a universal language for chemists, enabling precise calculations in chemical reactions, stoichiometry, and solution chemistry. By using moles, scientists can accurately predict the outcomes of chemical reactions and ensure the correct proportions of reactants are used in experiments and industrial applications.

Moving on to solutions, these are homogeneous mixtures consisting of two or more substances. In a solution, we have two main components: the solute and the solvent. The solute is the substance that dissolves, while the solvent is the substance that does the dissolving. Typically, the component present in larger quantity is considered the solvent.

To illustrate this concept, let's consider a common example: salt dissolved in water. In this case, salt (sodium chloride) is the solute, and water is the solvent. When salt is added to water, the ionic bonds between sodium and chloride ions break, and the ions become surrounded by water molecules. This process of dissolution creates a homogeneous mixture where the salt is evenly distributed throughout the water.

The particle model provides an excellent way to visualize the difference between solids and solutions. In a solid, such as a salt crystal, particles (atoms, molecules, or ions) are tightly packed in a fixed arrangement. They vibrate in place but do not move freely. When we dissolve this solid in a solvent, the particles separate and become dispersed throughout the liquid.

Using the particle model to depict a salt solution, we would see individual sodium and chloride ions surrounded by water molecules. These ions are no longer in a rigid structure but are free to move within the solution. This model helps explain why solutions, unlike solids, can flow and take the shape of their container.

The concept of solutions extends beyond just liquids dissolving solids. We can have gas-in-liquid solutions (like carbonated beverages), liquid-in-liquid solutions (such as alcohol in water), and even solid-in-solid solutions (as seen in some metal alloys). Each type of solution demonstrates the principle of one substance dispersing uniformly within another.

Understanding moles and solutions is essential for various applications in chemistry and everyday life. From preparing medications with precise concentrations to formulating cleaning products and cooking, these concepts are ubiquitous. In more advanced chemistry, knowledge of moles and solution chemistry is crucial for analyzing reaction rates, equilibrium constants, and thermodynamic properties.

In conclusion, moles provide a standardized way to count and measure particles in chemistry, while solutions represent a fundamental type of mixture where solutes are dissolved in solvents. The particle model helps us visualize these concepts, making it easier to understand the behavior of matter at the molecular level. Whether you're a student, a scientist, or simply curious about the world around you, grasping these concepts opens up a deeper appreciation for the intricate nature of chemical systems and their wide-ranging applications in our daily lives.

Concentration and Molarity

Concentration and molarity are fundamental concepts in chemistry that describe the amount of a substance dissolved in a solution. Understanding these concepts is crucial for various chemical calculations and applications. Let's delve into the definitions and relationships between concentration, molarity, moles, and volume.

Concentration refers to the amount of solute (the substance being dissolved) present in a given amount of solution. It can be expressed in various ways, but one of the most common and useful expressions is molarity. Molarity, denoted by the symbol M, is defined as the number of moles of solute per liter of solution. This relationship between moles, volume, and concentration is at the core of many chemical calculations.

The relationship between moles, volume, and concentration can be expressed using the concentration volume formula: C = n/V, where C represents the concentration (in mol/L or M), n is the number of moles of solute, and V is the volume of the solution in liters. This formula is essential for calculating concentration when given the number of moles and volume, or for determining the number of moles or volume when concentration is known.

Let's break down each component of the formula:

  • C (Concentration): Measured in mol/L or M (moles per liter or molarity)
  • n (Number of moles): Represents the amount of solute
  • V (Volume): Measured in liters (L)

The units used for concentration are typically mol/L or simply M (molarity). For example, a 1 M solution contains 1 mole of solute per liter of solution. It's important to note that molarity is temperature-dependent, as the volume of a solution can change with temperature.

To illustrate how to calculate concentration using the formula C = n/V, let's consider a few examples:

  1. Calculate the concentration of a solution containing 0.5 moles of NaCl dissolved in 2 liters of water: C = n/V = 0.5 mol / 2 L = 0.25 M
  2. Determine the concentration of a solution made by dissolving 40 grams of NaOH (molar mass 40 g/mol) in enough water to make 500 mL of solution: First, calculate the number of moles: n = 40 g / 40 g/mol = 1 mol Then, convert 500 mL to liters: 500 mL = 0.5 L Finally, apply the formula: C = 1 mol / 0.5 L = 2 M

Understanding the concentration volume formula and its applications is crucial for various chemical calculations, including preparing solutions of specific concentrations, diluting solutions, and determining the amount of solute needed to achieve a desired concentration. Mastering these concepts allows chemists and students to accurately prepare and analyze solutions in laboratory settings and real-world applications.

In conclusion, concentration and molarity are essential concepts in chemistry that describe the relationship between the amount of solute and the volume of solution. The concentration volume formula, C = n/V, provides a straightforward way to calculate and understand these relationships. By expressing concentration in mol/L or M, chemists can easily communicate and work with solution concentrations across various applications and experiments.

Calculating Moles from Volume and Concentration

Understanding how to calculate moles from volume and concentration is a fundamental skill in chemistry. This process is essential for various laboratory applications and chemical calculations. In this guide, we'll explore how to convert ml to moles and use concentration to determine the number of moles in a solution.

The Concentration Formula

The concentration of a solution is typically expressed in molarity (M), which is defined as moles of solute per liter of solution. The formula for concentration is:

C = n / V

Where:

  • C = concentration (in mol/L or M)
  • n = number of moles of solute
  • V = volume of solution (in liters)

Rearranging the Formula to Calculate Moles

To calculate moles from volume and concentration, we need to rearrange the concentration formula. The rearranged formula is:

n = C × V

This formula allows us to calculate the number of moles (n) when we know the concentration (C) and volume (V) of a solution.

Step-by-Step Example: Calculating Moles from mL and Concentration

Let's work through an example to demonstrate how to calculate moles from volume in milliliters (mL) and concentration.

Problem: Calculate the number of moles of NaCl in 250 mL of a 0.5 M NaCl solution.

Step 1: Identify the given information

  • Volume (V) = 250 mL
  • Concentration (C) = 0.5 M

Step 2: Convert mL to L

250 mL = 0.250 L

Step 3: Apply the formula n = C × V

n = 0.5 mol/L × 0.250 L = 0.125 mol

Answer: There are 0.125 moles of NaCl in 250 mL of a 0.5 M NaCl solution.

Practice Problems

To reinforce your understanding of how to calculate moles from volume and concentration, try these practice problems for moles calculation:

  1. Calculate the number of moles of H2SO4 in 100 mL of a 0.25 M H2SO4 solution.
  2. How many moles of KOH are present in 500 mL of a 1.5 M KOH solution?
  3. Determine the number of moles of glucose (C6H12O6) in 75 mL of a 0.8 M glucose solution.

Tips for Converting Volume to Moles

  • Always convert volume to liters before using the formula.
  • Pay attention to the units of concentration (usually mol/L or M).
  • Double-check your calculations to ensure accuracy.
  • Practice with various scenarios to become proficient in moles calculations.

Applications of Moles Calculations

Understanding how to calculate moles from volume and concentration is crucial in many areas of chemistry and related fields:

  • Preparing solutions for laboratory experiments
  • Analyzing chemical reactions and stoichiometry
  • Determining the amount of substance in pharmaceutical formulations

Converting Between Moles, Volume, and Concentration

Understanding the relationship between moles, volume, and concentration is crucial in chemistry and laboratory work. The formula C = n/V is a fundamental equation that connects these three important variables. In this formula, C represents concentration (usually in mol/L), n stands for the number of moles, and V is the volume (typically in liters). This equation allows us to convert between moles, volume, and concentration efficiently.

To convert from moles to volume when given concentration, we can rearrange the formula to V = n/C. This is particularly useful when we need to determine the volume of a solution with a known number of moles and concentration. For example, if we have 0.5 moles of a solute and want to prepare a 0.2 M solution, we can calculate the required volume as V = 0.5 mol / 0.2 mol/L = 2.5 L.

Conversely, to find moles from concentration and volume, we use the original formula n = C × V. This is handy when we need to determine the amount of solute in a given volume of solution with a known concentration. For instance, if we have 3 L of a 0.5 M solution, we can calculate the number of moles as n = 0.5 mol/L × 3 L = 1.5 mol.

The volume concentration formula is particularly useful in laboratory settings. Chemists often need to prepare solutions of specific concentrations, and this formula helps them determine the correct amounts of solutes and solvents to use. For example, when preparing buffer solutions or standardized reagents, precise calculations using this formula ensure accuracy in experiments and analyses.

Real-world applications of these calculations extend beyond the laboratory. In environmental science, researchers use these conversions to analyze pollutant concentrations in water bodies. The pharmaceutical industry relies on these calculations for drug formulation and dosage determination. In industrial chemical reactions, engineers use these principles to control chemical reactions and optimize product yields.

When working with gases, the ideal gas law applications (PV = nRT) comes into play, which incorporates pressure and temperature alongside moles and volume. This expands the applicability of mole-volume-concentration relationships to gaseous systems, crucial in atmospheric chemistry and industrial chemical reactions.

In titrations, a common analytical technique, the relationship between moles, volume, and concentration is essential for determining unknown concentrations of solutions. By knowing the concentration and volume of a standard solution, chemists can calculate the moles of the analyte in a sample.

Understanding how to convert between moles, volume, and concentration is also vital in stoichiometry calculations. These calculations are the foundation of quantitative chemistry, allowing scientists to predict the amounts of reactants needed or products formed in chemical reactions.

In summary, mastering the conversion between moles, volume, and concentration using the C = n/V formula is indispensable in chemistry. Whether you're finding volume from moles and concentration, calculating moles from concentration and volume, or applying the volume concentration formula, these skills are essential for accurate and efficient chemical analysis and preparation. The wide-ranging applications of these principles in laboratory work, environmental studies, pharmaceuticals, and ideal gas law applications underscore their importance in both academic and practical settings.

Introduction to Titration

Titration is a fundamental analytical technique in chemistry used to determine the unknown concentration of a chemical substance. This method is particularly useful in quantitative analysis and is widely employed in various fields, including environmental science, pharmaceuticals, and food industry. At its core, titration involves the controlled addition of a solution with a known concentration (the titrant) to a solution with an unknown concentration (the analyte) until a chemical reaction between the two is complete.

The basic principle of acid-base titration, one of the most common types, relies on the neutralization reaction between an acid and a base. In this process, a standard solution of known concentration (usually a strong acid or base) is gradually added to a measured volume of the solution with unknown concentration. As the titrant is added, it reacts with the analyte, causing a change in the solution's properties, such as pH.

Precise measurements are crucial in titration to ensure accurate results. This includes using calibrated glassware, such as burettes and pipettes, to measure volumes accurately. The concentration of the standard solution must be known with high precision, and the volume of the analyte must be measured carefully. Even small errors in measurement can lead to significant inaccuracies in the final calculation of the unknown concentration.

A key concept in titration is the equivalence point, which is the point at which the amount of titrant added is chemically equivalent to the amount of analyte present. At this point, the reaction between the titrant and analyte is complete. In acid-base titration, the equivalence point occurs when the number of moles of acid equals the number of moles of base, resulting in a neutral solution (pH 7 for strong acid-strong base titrations).

Identifying the equivalence point is crucial for accurate titration results. This is often done using indicators that change color at or near the equivalence point. For instance, phenolphthalein, a common indicator in acid-base titrations, turns from colorless to pink when the solution becomes slightly basic, signaling the end of the titration.

To illustrate how titration calculations work, let's consider a simple example. Suppose we want to determine the concentration of an unknown hydrochloric acid (HCl) solution. We might use a standardized sodium hydroxide (NaOH) solution with a known concentration of 0.1 M as our titrant. If it takes 25.0 mL of this NaOH solution to neutralize 20.0 mL of the unknown HCl solution, we can calculate the HCl concentration as follows:

1. Calculate the moles of NaOH used: 0.1 M × 0.025 L = 0.0025 moles NaOH

2. Since the reaction is 1:1 (HCl + NaOH NaCl + H2O), the moles of HCl must equal the moles of NaOH.

3. Calculate the concentration of HCl: 0.0025 moles ÷ 0.020 L = 0.125 M HCl

This example demonstrates how titration allows us to determine unknown concentrations through precise measurements and stoichiometric calculations. The accuracy of this method makes it invaluable in various analytical applications, from quality control in manufacturing to environmental monitoring and medical diagnostics.

Conclusion

In this article, we've explored the fundamental concepts of moles and concentration in chemistry. Understanding these key principles is crucial for success in chemical calculations and analysis. We've covered the definition of a mole, its relationship to Avogadro's number, and how it connects to concentration measurements. The importance of mastering mole calculations cannot be overstated, as they form the basis for more advanced chemistry topics. We encourage readers to practice these calculations regularly and refer back to the introductory video for visual explanations. Remember, proficiency in moles and concentration opens doors to exploring more complex chemical concepts. For further study, consider delving into topics such as stoichiometry, equilibrium, and reaction kinetics. By building a strong foundation in these areas, you'll be well-equipped to tackle more advanced chemistry challenges and deepen your understanding of the molecular world around us.

Expanding our Moles Calculations

Recap - Moles, Molar Volume, and Concentration

Step 1: Introduction to Moles and Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves using a chemical equation to determine the amount of substances required or produced in a chemical reaction. The key to stoichiometry is the mole, which represents a very large number of atoms or molecules, approximately 600 billion trillion. This large number allows chemists to work with quantities of substances that are practical to measure in the lab.

In stoichiometry, we often need to calculate the number of moles of a substance. This can be done using the molar mass for solids, the volume of gases, or the concentration of solutions. Each of these methods provides a way to measure the amount of a substance in different states of matter.

Step 2: Understanding Molarity and Concentration

Molarity, also known as molar concentration, is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. Molarity is an important concept because it allows chemists to quantify the concentration of a solution, which is crucial for many chemical reactions and processes.

To calculate molarity, you need to know the number of moles of the solute and the volume of the solution in liters. The formula for molarity (M) is:

M = moles of solute / liters of solution

This formula allows you to determine the concentration of a solution, which can then be used in various calculations involving chemical reactions in solutions.

Step 3: Calculating Moles in Solutions

When dealing with solutions, it is often necessary to calculate the number of moles of a solute. This can be done using the molarity of the solution and the volume of the solution. The formula to calculate the number of moles (n) is:

n = Molarity (M) × Volume (L)

By using this formula, you can determine the number of moles of a solute in a given volume of solution. This is particularly useful in titration experiments, where the concentration of one solution is used to determine the concentration of another.

Step 4: Applying Moles Calculations in Titrations

Titration is a common laboratory technique used to determine the concentration of a solution. In a titration, a solution of known concentration (the titrant) is added to a solution of unknown concentration (the analyte) until the reaction between the two is complete. The point at which the reaction is complete is known as the equivalence point.

To calculate the concentration of the analyte, you need to know the volume and concentration of the titrant, as well as the volume of the analyte. The relationship between the moles of the titrant and the moles of the analyte can be used to determine the concentration of the analyte.

The formula for this calculation is:

M1V1 = M2V2

Where M1 and V1 are the molarity and volume of the titrant, and M2 and V2 are the molarity and volume of the analyte. By rearranging this formula, you can solve for the unknown concentration of the analyte.

Step 5: Recap and Summary

In this section, we have expanded our understanding of moles calculations by exploring the concepts of molarity and concentration. We have learned how to calculate the molarity of a solution, how to determine the number of moles in a solution, and how to apply these calculations in titration experiments.

Understanding these concepts is crucial for accurately measuring and working with chemical substances in various states of matter. By mastering these calculations, you can confidently perform stoichiometric calculations and determine the amounts of substances involved in chemical reactions.

FAQs

Here are some frequently asked questions about moles and concentration:

1. How do you convert mL to moles?

To convert mL to moles, you need to know the concentration of the solution. Use the formula: moles = (volume in L) × (concentration in mol/L). First, convert mL to L by dividing by 1000, then multiply by the concentration.

2. What is the relationship between concentration and molar concentration?

Molar concentration is a specific type of concentration that expresses the number of moles of solute per liter of solution. It's typically measured in mol/L or M (molarity). Other concentration units can be converted to molar concentration if the molecular weight of the solute is known.

3. How do you calculate the number of moles from volume?

To calculate moles from volume, use the formula: n = C × V, where n is the number of moles, C is the concentration in mol/L, and V is the volume in liters. Make sure to convert the volume to liters if it's given in mL.

4. What is the concentration formula c1v1 = c2v2?

This formula is used for dilution calculations. It states that the product of the initial concentration (c1) and volume (v1) equals the product of the final concentration (c2) and volume (v2). It's useful for determining the concentration or volume after diluting a solution.

5. How do you convert percentage concentration to moles?

To convert percentage concentration to moles, first convert the percentage to a decimal (e.g., 5% = 0.05). For w/v percentages, this decimal represents grams per mL. Multiply by the total volume in mL to get grams, then divide by the molecular weight to get moles. For v/v percentages, additional density information may be needed.

Prerequisite Topics

Understanding moles and molar concentration is a fundamental concept in chemistry that builds upon several key prerequisite topics. While there are no specific prerequisite topics provided for this article, it's important to recognize that a strong foundation in basic chemistry principles is essential for grasping these concepts effectively.

Moles and molar concentration are central to many chemical calculations and processes. To fully comprehend these ideas, students should have a solid understanding of atomic structure, chemical formulas, and basic mathematical skills. These foundational concepts provide the necessary framework for exploring the world of moles and molar concentration.

Atomic structure knowledge helps students understand how atoms combine to form molecules and compounds, which is crucial when dealing with moles. Familiarity with chemical formulas allows for the accurate representation of substances and their relative quantities in chemical reactions. Additionally, basic mathematical skills, including algebra and unit conversions, are indispensable for performing mole-related calculations and determining molar concentrations.

The concept of moles serves as a bridge between the microscopic world of atoms and molecules and the macroscopic world of measurable quantities. It allows chemists to relate the number of particles in a substance to its mass and volume. This relationship is fundamental in stoichiometry, solution chemistry, and many other areas of chemical analysis.

Molar concentration, often expressed as molarity, is a measure of the amount of solute dissolved in a given volume of solution. This concept is crucial for understanding solution properties, reaction rates, and equilibrium processes. It builds upon the idea of moles and introduces the importance of volume in chemical calculations.

By mastering these prerequisite topics, students can more easily grasp the intricacies of moles and molar concentration. They will be better equipped to solve complex chemical problems, predict reaction outcomes, and understand the behavior of substances in various chemical environments.

As students progress in their chemistry studies, they will find that moles and molar concentration are recurring themes that connect various aspects of the subject. These concepts are essential for advanced topics such as acid-base chemistry, thermodynamics, and chemical kinetics. Therefore, a strong foundation in the prerequisites will not only aid in understanding moles and molar concentration but also pave the way for success in more advanced chemistry courses.

In conclusion, while specific prerequisite topics were not provided, it's clear that a solid grounding in basic chemistry principles is vital for mastering moles and molar concentration. Students should focus on strengthening their understanding of atomic structure, chemical formulas, and mathematical skills to build a robust foundation for these important chemical concepts.

In this lesson, we will learn:
  • To relate the mole concept to amount of substance in aqueous solutions.
  • To calculate the molarity of aqueous substances.
  • To calculate amounts of substance from titration problems.

Notes:

  • We saw in Moles, mass and gas calculations that for practical reasons, gas is usually measured in volume as opposed to mass which is used for solids. When we fix (keep constant) the temperature and pressure, at STP or RTP, the volume that one mole of an ideal gas occupies is also fixed, or constant.
    • At standard temperature and pressure (STP) it is 22.4 litres per mole (L mol-1).
    • At room temperature and pressure (RTP) it is 24 litres per mole (L mol-1).

    Like gases, when it comes to aqueous substances (something dissolved in water), mass and volume are not used. Instead, concentration is used.

  • Concentration effectively means “how much stuff in how much space?” which is why in chemistry concentration is measured in moles per litre, or moles per cubic decimetre (the value is equal).
    Sometimes the term molarity is used. Molarity means concentration, for chemists – molarity is the number of moles of a chemical per amount of volume.
    • Units of concentration are abbreviated “M”. It means moles per litre, written mol / L or mol L-1 or moles per cubic decimeters, written mol/dm3 or mol dm-3.
    • Square brackets, e.g. [HCl] are used to show the concentration of a chemical. For example, you could write [HCl] = 0.1 mol dm-3 which means there is 0.1 mole of HCl per 1 litre of this solution.

    You can use c=c = nV\large \frac{n}{V} to find concentration, where n = number of moles and VV is volume (in liters, L, or cubic decimeters, written dm3). You can then re-arrange for n=c∗vn = c * v.

  • Because concentration is measured in one unit per another unit (moles per litre), using the conversion factor method will not cancel your units like the examples with mass or volume. It is first easiest to separate the two unit conversions, combining them at the end.
    • For example, find the concentration of a solution of 54g of solid NaOH pellets being dissolved in 1500 mL of water.
      Our target units are moles per litre: we will find moles using the mass and molar mass first, then find litres using the mL value given and then we will divide the moles by litres:

    54 g NaOH ∗ g \, NaOH \,* \, 1 mol NaOH40 g NaOH\large \frac{1 \, mol \, NaOH} {40 \, g \, NaOH} = = 1.35 mol NaOHmol \, NaOH

    1500 mL NaOH ∗ mL \, NaOH \,* \, 1 L NaOH1000 mL NaOH\large \frac{1 \, L \, NaOH} {1000 \, mL \, NaOH} = = 1.57 L NaOHL \, NaOH

    Using C=n/vC = n/v we can now find the concentration by dividing moles by volume:

    1.35 mol NaOH1.5 L NaOH\large \frac{1.35 \, mol \, NaOH} {1.5 \, L \, NaOH} = = 0.9 M NaOHM \, NaOH

  • Using concentration to find the number of moles is very useful for knowing the amounts of substance in titration experiments.
    A titration is an experiment used to find out the unknown concentration of an acid by reacting it with a base of known concentration, or vice versa (unknown base with known acid). We call the solution of known concentration a standard solution. The concentration of this is known precisely.
    For an acid of known concentration reacting with a base of unknown concentration:
    • A titration experiment runs until the number of moles of acid equals the number of moles of base, which is found by a colour change using an indicator.
    • The known acid concentration and volume of acid is used to find the number of moles of acid, which is used to find the number of moles of the base of unknown concentration.
      Look for the molar ratio from the chemical equation, for example if acid A and base B react in the equation 2A + B   \, →  \, C + H2O to make salt C, there is a 2:1 molar ratio. The moles of A will be twice the moles of B.
    • You use the number of moles found and the volume of the unknown concentration to find the concentration of the unknown substance.

    In the exercises of this lesson we will practice worked calculations using data from titration experiments, using the equation for concentration and unit conversions shown above. The detailed process of titration experiments is looked at in Acid-base titration

  • Be careful with units of concentration, converting units if you need to. Volume is often given in mL but concentration is measured in moles per litre or moles per cubic decimeter, which has the same value. Dividing by 1000 converts from mL to L.