Acid-Base Titration: Mastering the Math Behind Chemistry
Dive into the world of acid-base titration math! Learn to calculate concentrations, find equivalence points, and apply your skills to real-world scenarios. Our expert guidance makes complex concepts clear.

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Now Playing:Acid base titration – Example 0a
Intros
  1. What is titration?
  2. What is titration?
    How titration works
  3. What is titration?
    Finding unknown concentration: titration calculations.
Examples
  1. Determine the unknown concentration from acid-base titration results:
    A solution of KOH of unknown concentration was titrated by 0.3 M hydrochloric acid. The results below show the volume of HCl required to titrate 25cm3 of the KOH solution:

    Run:

    Titre volume (cm3)

    1

    18.60

    2

    18.35

    3

    18.40

      1. Which two of these readings are concordant and should be used to find the average titre?
      2. Calculate the average titre for this experiment.

    1. Use your average titre to find the concentration of the potassium hydroxide (KOH) solution.

Introduction to acid-base theory
Notes

In this lesson, we will learn:

  • Why titrations are performed and their purpose as an experiment.
  • How to use reaction stoichiometry to find moles and unknown concentrations from titrations.
  • How to use titrations and mole calculations to find percentage purity of a sample.

Notes:

  • A titration is a lab experiment used to find the unknown concentration of an acid or base. We can use an acid or base sample with a known concentration (let’s call the ‘known’ chemical A) in a chemical reaction with a sample of a base or acid of unknown concentration that we’re investigating (let’s call the ‘unknown’ chemical B).
    Chemicals A and B will react in proportional amounts (moles) to each other, so the amounts of A we used, in the end, tells us the amounts of B that’s present.
    • Titration experiments work because volume AND concentration of chemical A are known, so we can calculate moles of A, which is related to moles of B (in their reaction), which is used with its volume to calculate A’s unknown concentration! Finding this works in a sequence as shown in the table below:

  • The practical experiment goes like this:
    • Known chemical A is gradually added to a precise volume of unknown B until the stoichiometric point is reached. The stoichiometric point is when the ratio of moles of A to B is the same as in the balanced chemical equation. This point is clear from a large change in pH (and color change with indicator added) because neutralization occurs.
      The molar ratio in the reaction equation (the stoichiometry) and the volume of A that was needed to neutralize B is then used to find the number of moles of B. Now that moles and volume of ‘unknown’ B are known, the concentration of the unknown chemical B sample can be calculated.

  • Titrations are excellent for finding acid/base concentrations because:
      • Acids and bases are complementary in their chemistry so there are lots of ideal ‘partner’ chemicals for them.
      • It is easy to measure pH and pH change of a mixture, and the pH can be used to find the concentration and moles of H3O+ or OH- in that same mixture. This is perfect for chemicals that produce these ions, like acids and bases!

  • The most important factor in a successful titration is precision. The titration needs to stop precisely at the stoichiometric point (AKA equivalence point). To make sure this is accurate and precise, a few steps are taken in the practical method:
    • A chemical called an indicator is added to the mixture. This chemical indicates a change in pH by a clear color change, and the equivalence point is exactly when the titration mixture changes color permanently. There are many different indicators for different pH ranges and combinations of weak/strong acids reacting with weak/strong bases. They are very sensitive to pH change – the addition of one extra drop of acid or base might cause a permanent color change.
    • A trial run is performed on the very first titration experiment. This is a ‘rough run’ to establish roughly the volume of chemical A needed to reach the equivalence point. This is done to save time and still allow chemists to be precise in the approximate range where the color change – and equivalence point – is found.
    • The final amounts of the known chemical A are added dropwise to get as precise a reading as possible. A single titration run with a volume reading obtained is called a ‘titer’.
    • The precise runs – titers – are repeated until two readings fall within 0.05cm3 of each other. An average titer is then taken from these two readings.

  • Once an average titer has been obtained, this is taken to be the volume where there is an equal number of moles of both acid and base.
    Remember that your titer volumes will be in milliliters or cm3 but concentration is measured in moles per LITER, or dm3. You must convert the volume to liters! Dividing milliliters by 1000 will give you a value in liters.
    The number of moles of chemical B can be calculated because its concentration and volume are both known:

    concentration  (M)=moles  (mol)Volume  (L)concentration \; (M) = \frac{moles\; (mol)}{Volume \; (L)}

    Rearrange to give:

    moles  (mol)=concentration  (M)    Volume  (L)moles \; (mol) = concentration \;(M)\; * \; Volume \; (L)

    With this moles of B value obtained, recall the reaction stoichiometry:

    aA + bB   \;  \; cC + dD

    The molar ratio is important for finding moles of B. The equivalence point occurs when b moles of A is equal to a moles of B. Use this to find the number of moles of unknown chemical B in the titration mixture at equivalence point.

  • Once the number of moles of chemical B is found, you have moles and volume:

    concentration  (M)=moles  (mol)Volume  (L)concentration \; (M) = \frac{moles\; (mol)}{Volume \; (L)}

    Now the unknown concentration can be calculated!

  • Titrations involving partial neutralization can also be performed; this is where polyprotic acids (acids with more than one proton to donate) donate a proton one by one, so no one step completely neutralizes the acid. Take the example with phosphoric acid, H3PO4:

    Reaction 1: H3PO4 + OH-   \;  \; H2PO4- + H2O

    Reaction 2: H3PO4 + 2OH-   \;  \; HPO42- + 2H2O

    Reaction 3: H3PO4 + 3OH-   \;  \; PO43- + 3H2O

    If you are asked to do a calculation with partial neutralization, you might need to find out the acid:base molar ratio in the reaction. For this, you’ll be given the volume and concentration of both acid and base so that moles of both can be found – then divide the larger number of moles by the smaller to find the ratio between them! To find the moles, the rearranged equation for concentration can still be used:

    moles  (mol)=concentration  (M)    Volume  (L)moles \; (mol) = concentration \;(M)\; * \; Volume \; (L)

    This is sometimes written as: 

    n = C * V \quad Where \, n = moles, C = concentration (M), V = volume (L or dm3)

  • Titration results can be used to find the percentage purity of a sample as well. This is done in the following method:
    • Find the number of moles of the acid/base you added to the solvent to create the solution. For example, a 1.13g sample of NaOH was used. Find the number of moles of this by dividing the mass used by molar mass:

      moles=mass  (g)molar  mass  (gmol1)moles = \frac{mass \; (g)}{molar \; mass\; (gmol^{-1})}

      moles=1.1340=0.028molmoles = \frac{1.13}{40} = 0.028 mol

      Find the expected concentration by dividing the volume of the solution made by this number of moles.
      For example, the NaOH used was diluted to 250 mL (0.25 L):

      concentration  (M)=moles  (mol)volume  (L)concentration\; (M) = \frac{moles \; (mol)}{volume\; (L)}

      concentration  (M)=0.028  mol0.25(L)=0.112  Mconcentration\; (M) =\frac{0.028\; mol}{0.25 (L)} = 0.112\; M

      Record your expected concentration.

    • Perform the titration and find the concentration of your sample substance using your titer results with the method above. This is your actual concentration – if you have impurities in your sample, they won’t react but they still took up mass in your sample, so this actual concentration will be lower than the expected concentration.
      For example, a 25.00 mL sample of the NaOH solution was titrated with 27.90 mL (or 0.0279 L) of 0.097 M HCl:

      moles  (mol)  =0.097M0.0279L=0.0027molmoles \; (mol) \; = 0.097M * 0.0279L = 0.0027 mol

      HCl and NaOH react in a 1:1 ratio, so 0.0027 mol HCl = mol NaOH:

      concentration  (M)=0.0027  mol0.025  L=0.108Mconcentration \; (M) = \frac{0.0027\; mol}{0.025\; L} = 0.108M

    • Use these two values to calculate percentage purity of your sample with the equation:

      %  purity=actual  concentrationexpected  concentration100=1.08  M1.12  M100=96.4%\% \; purity = \frac{actual \; concentration}{expected \; concentration} * 100= \frac{1.08\;M}{1.12\;M} * 100 = 96.4 \%
Concept

Introduction to Acid-Base Titration

Acid-base titration is a fundamental chemistry experiment that plays a crucial role in quantitative analysis. This precise technique allows chemists to determine the concentration of an unknown acid or base solution by neutralizing it with a standard solution of known concentration. The importance of titration in chemistry cannot be overstated, as it forms the basis for numerous analytical procedures in research, industry, and environmental monitoring. Our introductory video provides a visual demonstration of this essential process. In this article, we'll delve into the basics of acid-base titration, exploring the underlying principles and step-by-step procedures. We'll also cover the calculations involved in interpreting titration results, including how to determine equivalence points and calculate unknown concentrations. Finally, we'll examine the wide-ranging applications of titration in various fields, from pharmaceutical quality control to food and beverage analysis, highlighting its significance in real-world scenarios.

FAQs
  1. How do you calculate titration?

    To calculate titration results, use the equation c1v1 = c2v2, where c is concentration and v is volume. Knowing the concentration and volume of the titrant, along with the volume of the analyte, you can determine the unknown concentration. For example, if 20 mL of 0.1 M NaOH neutralizes 25 mL of HCl, you'd calculate: (0.1 M)(20 mL) = (x M)(25 mL), solving for x to find the HCl concentration.

  2. What is the definition of titration in math?

    In mathematical terms, titration is a process of determining the concentration of a solution by reacting it with a known volume of another solution of known concentration. It's represented by the equation c1v1 = c2v2, where c1 and v1 are the concentration and volume of the known solution, and c2 and v2 are those of the unknown solution.

  3. What is titration for dummies?

    Titration for beginners can be explained as a method of finding out how much of a substance is in a solution by slowly adding another substance that reacts with it until the reaction is complete. It's like filling a bathtub to a specific level - you add water slowly until it reaches the right mark, then you know how much water you've added.

  4. Is titration always 1 to 1?

    No, titration is not always 1 to 1. The ratio depends on the stoichiometry of the reaction. While many acid-base titrations are 1:1, some reactions, like the titration of sulfuric acid (H2SO4) with sodium hydroxide (NaOH), have a 1:2 ratio. Always refer to the balanced chemical equation to determine the correct ratio for calculations.

  5. How do you get the best titration results?

    To get the best titration results: 1) Use clean, calibrated glassware. 2) Rinse the burette with the titrant before filling. 3) Remove air bubbles from the burette tip. 4) Add titrant slowly, especially near the endpoint. 5) Use a white background to see color changes clearly. 6) Perform at least three trials for consistency. 7) Calculate the average of your results. 8) Use a suitable indicator or pH meter for accurate endpoint detection.

Prerequisites

Understanding acid-base titration is crucial in chemistry, but to truly grasp this concept, it's essential to have a solid foundation in certain prerequisite topics. Two key areas that significantly contribute to comprehending acid-base titration are balancing chemical equations and the acid dissociation constant.

Let's start with the importance of balanced chemical equations. In acid-base titration, we're dealing with reactions between acids and bases. To accurately predict and analyze these reactions, it's crucial to work with properly balanced equations. Balancing ensures that we account for all atoms and charges on both sides of the equation, which is fundamental for stoichiometric calculations in titration. Without this skill, students might struggle to determine the correct amounts of reactants or products, leading to inaccurate titration results.

Moreover, understanding how to balance equations helps in visualizing the neutralization process that occurs during titration. It allows students to see how protons are transferred between the acid and base, and how this affects the overall reaction. This visualization is key to grasping the concept of equivalence point in titration.

Equally important is the concept of the acid dissociation constant. This fundamental principle is crucial for understanding the strength of acids and bases, which directly impacts titration procedures and calculations. The acid dissociation constant, often denoted as Ka, provides information about how readily an acid gives up its protons in solution. This knowledge is essential for predicting the behavior of acids and bases during titration and for interpreting titration curves.

Furthermore, familiarity with polyprotic acid dissociation is particularly relevant in more complex titrations. Many real-world titrations involve polyprotic acids, which have multiple dissociation steps. Understanding how these acids behave at different stages of dissociation is crucial for accurately interpreting titration data and identifying multiple equivalence points.

By mastering these prerequisite topics, students build a strong foundation for understanding acid-base titration. They'll be better equipped to perform titrations, interpret results, and solve related problems. The ability to balance equations ensures accurate stoichiometric calculations, while knowledge of acid dissociation constants allows for a deeper understanding of acid-base behavior and strength. Together, these skills enable students to approach acid-base titration with confidence and precision, paving the way for success in more advanced chemistry concepts.