About this Video

• Video Title: KMnO4 Vs Oxalic acid Redox titration calculations by Seema Makhijani
• Channel: Seema Makhijani
• Speakers: Seema Makhijani
• Duration: 00:14:39

Introduction

This video explains the calculations involved in the redox titration of potassium permanganate (KMnO4) against oxalic acid. The instructor, Seema Makhijani, derives the necessary formula from the balanced chemical equation and demonstrates its application with an example problem, including the calculation of the molarity and strength of KMnO4, as well as percentage purity if required.

Key Takeaways

• Balanced Chemical Equation: The video meticulously balances the redox reaction between KMnO4 and oxalic acid, highlighting the importance of balancing oxygen and hydrogen atoms in an acidic medium. The final balanced equation is crucial for deriving the mole ratio.
• Mole Ratio and Formula Derivation: The mole ratio between KMnO4 and oxalic acid (2:5) is derived from the balanced equation and used to create a formula relating the molarity and volume of each reactant: 5 * M_KMnO4 * V_KMnO4 = 2 * M_{oxalic acid} * V_{oxalic acid}.
• Preparation of Oxalic Acid Solution: The video shows how to prepare a standard solution of oxalic acid with a known molarity (M/40) by calculating the required weight based on molarity, molar mass (including 2H2O), and desired volume (100ml).
• Titration Calculation and Strength: The video demonstrates how to calculate the molarity of KMnO4 using the derived formula and titration data (volumes and known molarity of oxalic acid). It further explains the calculation of the strength of KMnO4 using its molarity and molar mass (158 g/mol).
• Percentage Purity Calculation: The video explains how to calculate the percentage purity of KMnO4 if required, using the calculated strength and a given strength value (both in grams per liter).

The video begins with Seema Makhijani greeting the class and stating that she will be discussing the titration of KMnO4 versus oxalic acid, focusing on the calculation aspect. She emphasizes that the formulas used are derived from the reaction itself.

The first reaction discussed is the reduction of KMnO4 to Mn²⁺ in a highly acidic medium (using dilute sulfuric acid). She balances the equation step-by-step:

1. Oxygen Balancing: Four oxygen atoms on the left (KMnO4) are balanced by adding four water molecules to the right.

2. Hydrogen Balancing: Eight hydrogen atoms are added to the left (as 8H⁺) to balance the hydrogens in the water molecules.

3. Charge Balancing: To balance the charge (8+ on the left and 2+ on the right), five electrons are added to the left side. This results in a balanced half-reaction:

   2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O

Next, the oxidation of oxalic acid (C₂H₂O₄) to CO₂ is balanced:

1. Carbon Balancing: Two carbon atoms are present on the left, so a coefficient of 2 is added to CO₂ on the right.

2. Oxygen Balancing: Oxygen is already balanced (4 on each side).

3. Hydrogen Balancing: No hydrogen balancing is needed.

4. Charge Balancing: Two electrons are added to the right to balance the charge, resulting in the balanced half-reaction:

   C₂H₂O₄ → 2CO₂ + 2H⁺ + 2e⁻

To obtain the overall balanced equation, the first half-reaction is multiplied by 2 and the second by 5 to equalize the number of electrons (10 electrons each). The equations are then added, and the electrons cancel out. The resulting balanced equation is:

2MnO₄⁻ + 16H⁺ + 5C₂H₂O₄ → 2Mn²⁺ + 8H₂O + 10CO₂

She notes that this titration occurs at 60-70°C, and the heating helps to remove CO₂ gas, driving the reaction forward. Initial heating is sufficient; continuous heating isn't needed because the reaction acts as an autocatalyst.

The formula for the mole ratio is derived from the balanced equation:

Moles of KMnO₄ / Moles of oxalic acid = 2/5

This leads to the formula:

5 * moles of KMnO₄ = 2 * moles of oxalic acid

Substituting molarity and volume for moles (moles = molarity * volume):

5 * M_KMnO₄ * V_KMnO₄ = 2 * M_{oxalic acid} * V_{oxalic acid}

This formula is the basis for all KMnO₄ vs. oxalic acid titration calculations.

An example is given where a standard solution of oxalic acid needs to be prepared. The goal is to make 100ml of an M/40 solution. The calculation involves using the formula:

Molarity = (weight of oxalic acid / molar mass of oxalic acid) / volume (in liters)

The molar mass of oxalic acid dihydrate (C₂H₂O₄·2H₂O) is 126 g/mol (this is emphasized as a common point of error). Substituting the known values, the required weight of oxalic acid is calculated to be 0.315g. The preparation procedure involves weighing 0.315g of oxalic acid, adding it to a 100ml volumetric flask using a funnel, and then adding water to the 100ml mark.

The calculation of the molarity of KMnO₄ involves substituting the values from the titration into the derived formula. At least three concordant readings are needed for accuracy. The molarity of KMnO₄ is then calculated and expressed to at least three decimal places.

Finally, the calculation of the strength of KMnO₄ is explained:

Strength = Molarity of KMnO₄ * Molar mass of KMnO₄

The molar mass of KMnO₄ is 158 g/mol. The units of strength are grams per liter.

If percentage purity is requested, the formula is:

Percentage purity = (Calculated strength / Given strength) * 100

The given strength would be provided in the question, and both strengths must be in grams per liter. The final answer is a percentage without units. The video concludes with encouragement for the students' board exams.

There is no word "supsup" in the transcript. There may be a misspelling or misunderstanding.

The correct formula derived in the video for the titration of KMnO₄ against oxalic acid is:

5 * M_KMnO₄ * V_KMnO₄ = 2 * M_{oxalic acid} * V_{oxalic acid}

Where:

• M_KMnO₄ = Molarity of KMnO₄ solution
• V_KMnO₄ = Volume of KMnO₄ solution used in the titration
• M_{oxalic acid} = Molarity of the oxalic acid solution
• V_{oxalic acid} = Volume of oxalic acid solution used in the titration

The numbers 5 and 2 come from the stoichiometric coefficients in the balanced chemical equation for the reaction.

That's the HTML tag for subscript. It's used to make text appear smaller and slightly lower than the baseline, like this: H₂O (the 2 is a subscript). It's not something that can be "corrected" on its own; it needs to be used within a larger text string to have meaning. Do you have a specific example of where you're seeing this tag and what you'd like me to do with it?

In the context of chemical formulas and equations (like the one you provided earlier), <sup> is the HTML tag for a superscript. Superscripts are used to show characters or numbers that are slightly smaller and raised above the baseline text. In chemical formulas, they are commonly used to indicate:

• Charges on ions: For example, the positive charge on Mn²⁺ (Manganese(II) ion) indicates a +2 charge.
• Isotopes: The mass number of an isotope is often written as a superscript to the left of the element symbol (e.g., ²³⁵U).

If you are using a plain text editor, the <sup> tag will not render properly; you will see the raw tag instead of the formatted superscript. To display superscripts in plain text, you generally just use the caret symbol ^. For example, Mn^2+ would visually represent the same information as Mn²⁺.

In short, <sup> is used for formatting a superscript, primarily for expressing charges or isotope mass numbers in chemistry.

The video covers the calculations needed for a potassium permanganate (KMnO4) titration against oxalic acid. The instructor, Seema Makhijani, starts by deriving the necessary formula from the balanced chemical equation. She emphasizes that understanding the reaction is key to accurate calculations.

The reaction shows KMnO4 changing to Mn2+ in a highly acidic solution (using dilute sulfuric acid). She carefully balances the equation, explaining each step. Similarly, the oxidation of oxalic acid to CO2 is balanced step-by-step. The two balanced half-reactions are then combined to get the overall balanced equation.

The balanced equation reveals the mole ratio of KMnO4 to oxalic acid, which is then used to create a formula linking the molarity and volume of each reactant. This formula is the foundation for all the calculations in this type of titration.

The video then shows how to prepare a standard oxalic acid solution of known concentration. The instructor explains the calculation, including the molar mass of the oxalic acid (taking into account water molecules), and clearly shows how to prepare the solution.

The titration calculations are demonstrated. The video uses the derived formula and real titration data to calculate the molarity of the KMnO4 solution. It emphasizes the need for several consistent titration readings to ensure accuracy.

The process of calculating the strength of the KMnO4 solution is also explained. This calculation uses the molarity of KMnO4 and its molar mass. The units of strength are grams per liter. The instructor also clearly explains how to calculate the percentage purity of the KMnO4 solution if needed, which uses the calculated strength and a given strength value. The video concludes by wishing the students success in their exams.

Seema Makhijani begins the video by introducing the topic: calculating the results of a redox titration between potassium permanganate (KMnO4) and oxalic acid. She emphasizes that understanding the underlying chemical reaction is crucial for deriving the correct formulas.

She starts by explaining the reduction half-reaction of KMnO4 in a strongly acidic environment (using dilute sulfuric acid). The process of balancing this equation is demonstrated step-by-step: oxygen is balanced by adding water, hydrogen is balanced by adding H+ ions, and finally, the charge is balanced by adding electrons. This results in a balanced half-reaction.

Next, she balances the oxidation half-reaction of oxalic acid, which is converted to carbon dioxide. This process is similarly explained step-by-step, ensuring the equation is balanced in terms of carbon, oxygen, hydrogen, and charge. This also yields a balanced half-reaction.

To obtain the overall balanced redox reaction, she multiplies each half-reaction by an appropriate factor to make the number of electrons equal in both. Then, she combines the half-reactions, and the electrons cancel out, giving the complete balanced equation. She mentions that the reaction is carried out at a temperature of 60-70°C, and explains that this heating helps drive the reaction forward by facilitating the removal of CO2 gas, which acts as a product. She notes that the reaction is autocatalytic, meaning once it starts, it continues without the need for continuous heating.

From the balanced equation, the mole ratio between KMnO4 and oxalic acid is determined. This ratio is directly used to create a formula that relates the molarity and volume of both reactants in the titration. This formula is central to all subsequent calculations.

She then provides a practical example: preparing a standard solution of oxalic acid. She guides through the calculation of the necessary weight of oxalic acid dihydrate needed to create a specific volume (100 ml) of a solution with a known molarity (M/40). The molar mass calculation for oxalic acid dihydrate is explicitly shown, emphasizing the importance of including the water molecules in the molar mass calculation. She shows the steps to prepare this solution in a volumetric flask.

The core calculation of the KMnO4 molarity is shown, using the previously derived formula and data obtained from a titration. The significance of obtaining multiple concordant readings is emphasized to improve accuracy and reliability.

The video continues by demonstrating how to calculate the strength of the KMnO4 solution, which involves using the calculated molarity and the molar mass of KMnO4. The units for strength are clarified as grams per liter.

Finally, she explains how to determine the percentage purity of the KMnO4, should the question require it. This calculation uses the calculated strength and a given (or known) strength value. Both values must be in grams per liter, and the final answer is a percentage, without any units.

The entire video provides a step-by-step approach to understanding and performing the calculations involved in KMnO4-oxalic acid titrations, starting from the basic principles of balancing chemical equations and leading to the practical applications of calculating molarity, strength, and percentage purity. The video concludes with encouraging words for the students.

The primary equation discussed and used throughout the video is the balanced chemical equation for the redox reaction between potassium permanganate (KMnO4) and oxalic acid (C₂H₂O₄) in an acidic medium. While the video breaks down the balancing process step-by-step, the final balanced equation is central to all calculations. The exact form might vary slightly depending on how the half-reactions are combined, but it will represent the same stoichiometry. One possible representation of the balanced equation is:

2KMnO₄ + 5C₂H₂O₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 10CO₂ + 8H₂O

Additionally, a crucial derived equation is used for calculating the molarity of KMnO₄ based on the titration results with oxalic acid:

5 * M_KMnO₄ * V_KMnO₄ = 2 * M_{oxalic acid} * V_{oxalic acid}

Where:

• M_KMnO₄ represents the molarity of the potassium permanganate solution.
• V_KMnO₄ represents the volume of potassium permanganate solution used.
• M_{oxalic acid} represents the molarity of the oxalic acid solution.
• V_{oxalic acid} represents the volume of oxalic acid solution used.

These two equations—the balanced chemical equation and the derived molarity equation—are the core equations presented in the video.

You are right to question that. My apologies; I am still under development and learning to perfectly adapt my responses to different contexts. While subscripts are essential in chemical formulas to represent the number of atoms, in a plain text format like this, they can be confusing and difficult to read. I should have presented the equations using a clearer method, like using parentheses or a different notation to convey the same information without relying on subscripts that won't render correctly for all users. I will strive to improve my ability to avoid such formatting issues in the future.

Seema Makhijani's video focuses on calculating the results of a potassium permanganate (KMnO4) titration against oxalic acid. She emphasizes that a thorough understanding of the underlying chemical reaction is crucial for accurate calculations.

The video begins by explaining the reduction half-reaction of KMnO4 in a strongly acidic solution (using dilute sulfuric acid). The balancing of this equation is shown step-by-step: oxygen is balanced using water, hydrogen is balanced using H+ ions, and charge is balanced using electrons. This process results in a balanced half-reaction.

Next, the oxidation half-reaction of oxalic acid (converted to carbon dioxide) is balanced in a similar step-by-step manner. The resulting balanced half-reaction is also shown.

To find the overall balanced equation, both half-reactions are multiplied by factors to equalize the number of electrons. These modified half-reactions are then added, with the electrons canceling out to yield the complete balanced equation. A possible representation of the balanced equation is:

2KMnO4 + 5C2H2O4 + 3H2SO4 --> K2SO4 + 2MnSO4 + 10CO2 + 8H2O

(Note: The numbers in this equation represent the stoichiometric coefficients and are not subscripts within the chemical formula). The reaction temperature (60-70°C) is mentioned, along with the explanation that this heating helps remove CO2 and thus drives the reaction forward. The autocatalytic nature of the reaction—once started, it continues without needing continuous heating—is also discussed.

The mole ratio between KMnO4 and oxalic acid is determined from the balanced equation. This ratio is used to create a formula connecting the molarity and volume of both reactants:

5 * (KMnO4 molarity) * (KMnO4 volume) = 2 * (oxalic acid molarity) * (oxalic acid volume)

This formula is key to all subsequent calculations.

A practical example follows: preparing a standard oxalic acid solution. The calculation of the required weight of oxalic acid dihydrate (taking water molecules into account for molar mass) to create a 100ml solution of a known molarity (M/40) is shown. The preparation steps using a volumetric flask are also described.

The core calculation—determining KMnO4 molarity using the formula above and titration data—is demonstrated. The importance of consistent titration readings is stressed for accuracy.

Calculating the strength of the KMnO4 solution is explained next. This involves using the calculated molarity and the molar mass of KMnO4. The units of strength (grams per liter) are clearly stated.

Finally, the calculation of KMnO4 percentage purity (if required) is shown. This calculation uses the calculated strength and a given strength value (both in grams per liter). The final answer is a percentage.

The video provides a comprehensive, step-by-step approach to the calculations involved in KMnO4-oxalic acid titrations, starting from balancing equations and progressing to the practical calculations of molarity, strength, and percentage purity.

No, not all carboxylic acids are reduced to CO2. While oxalic acid is readily reduced to CO2 under the specific conditions of the KMnO4 titration (strong acid and heat), other carboxylic acids may undergo different reduction reactions depending on the reducing agent and reaction conditions. The reduction products can vary greatly. Some might be reduced to aldehydes, alcohols, or other functional groups, not always completely to CO2.

The reduction products of carboxylic acids depend heavily on the reducing agent used and the reaction conditions (such as temperature, pH, and solvent). There isn't one single answer. Here are some possibilities:

• Aldehydes: Mild reducing agents like lithium aluminum hydride (LiAlH4) in controlled conditions can reduce carboxylic acids to aldehydes. However, aldehydes are often further reduced to alcohols if the conditions aren't carefully controlled.

• Alcohols: Stronger reducing agents like LiAlH4 (in excess) or sodium borohydride (NaBH4) typically reduce carboxylic acids all the way to primary alcohols.

• Other functional groups: Depending on the specific carboxylic acid and the reducing agent, other functional groups might be formed. For example, certain complex reducing agents might lead to the formation of esters or other derivatives.

• Decarboxylation: Some reactions might result in the removal of the carboxyl group (-COOH) completely, leaving behind a hydrocarbon chain. This is known as decarboxylation. It's not technically a reduction of the entire molecule, but rather a modification.

In short, the reduction product isn't solely determined by the carboxylic acid itself but is a complex interplay of factors. If you have a specific carboxylic acid and reducing agent in mind, a more precise answer can be given based on the known chemistry of that reaction.