4.79 | Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end
TLDRIn this educational video, the presenter explains how to determine the concentration of sulfuric acid in rainwater using a titration method. They use the formula m1V1 = m2V2, adapted for acids and bases, where molarity is multiplied by the number of H+ ions for the acid and OH- ions for the base. The process involves a 20 mL sample of acid rain requiring 1.7 mL of 0.0811 M NaOH to reach the endpoint. By applying the formula and solving for the unknown molarity of H2SO4, the video provides a clear demonstration of acid-base titration, resulting in a molarity of approximately 3.4 x 10^-3 M for sulfuric acid.
Takeaways
- 🔍 The script discusses a titration experiment involving a 20 mL sample of acid rain and a 0.0811 M sodium hydroxide (NaOH) solution.
- 🧪 The experiment aimed to determine the concentration of sulfuric acid (H2SO4) in the acid rain sample, assuming it is the cause of the acidity.
- 📚 It is mentioned that the balanced chemical equation is not provided but is not necessary for solving the problem.
- 🔬 The titration process involves adding a strong base (NaOH) to a strong acid (H2SO4) to form a salt and water, reaching an endpoint indicated by a color change.
- 📈 A shortcut formula is introduced for acid-base titrations, relating the molarity and volume of the acid to those of the base, accounting for the number of H+ and OH- ions.
- ⚖️ The formula used is \( 2 \times x \times 20 = 0.0811 \times 1.7 \), where \( x \) represents the molarity of H2SO4, with the factor of 2 accounting for the two H+ ions in H2SO4.
- 🧐 The volumes used in the formula do not need to be in liters and can be in mL, as long as the units are consistent across the equation.
- 🔢 The calculation simplifies to solving \( 40x = 0.13787 \), which is then divided by 40 to find the molarity of H2SO4.
- 📝 The final answer for the molarity of H2SO4 is approximately \( 3.4 \times 10^{-3} \) M, considering significant figures.
- 👨🏫 The script emphasizes the ease of using the formula for titration calculations compared to writing out the balanced chemical equation.
- 🌟 The video concludes with an encouragement for viewers to apply the formula to their own titration problems and to reach out with questions.
Q & A
What is the purpose of a titration in a chemistry lab?
-A titration is a laboratory technique used to determine the concentration of an unknown acid or base by reacting it with a solution of known concentration, the titrant.
What is the significance of the endpoint in a titration process?
-The endpoint is the point at which the titration is considered complete, usually indicated by a color change in the flask due to an indicator added to the solution.
What is the balanced chemical equation for the reaction between sodium hydroxide (NaOH) and sulfuric acid (H2SO4)?
-The balanced chemical equation is 2 NaOH + H2SO4 → Na2SO4 + 2 H2O. This shows that two moles of NaOH react with one mole of H2SO4 to form sodium sulfate and water.
Why is it important to memorize strong acids and bases for titration calculations?
-Memorizing strong acids and bases is important because they dissociate completely in water, which simplifies the calculations for titrations, as all the acid or base is available to react.
What is the shortcut formula used for acid-base titrations?
-The shortcut formula is m1V1 = m2V2, where m1 and V1 are the molarity and volume of the acid, and m2 and V2 are the molarity and volume of the base. For acids, multiply by the number of H+ ions, and for bases, multiply by the number of OH- ions.
How many moles of H+ ions does sulfuric acid (H2SO4) release when it dissociates in water?
-Sulfuric acid (H2SO4) releases two moles of H+ ions when it dissociates in water, as it is a diprotic acid.
What is the volume of the acid rain sample used in the titration described in the script?
-The volume of the acid rain sample used in the titration is 20 mL.
What is the molarity of the sodium hydroxide (NaOH) solution used to titrate the acid rain sample?
-The molarity of the sodium hydroxide (NaOH) solution used in the titration is 0.0811 M.
How much volume of NaOH solution was required to reach the endpoint in the titration?
-1.7 mL of the 0.0811 M NaOH solution was required to reach the endpoint in the titration.
What was the calculated molarity of sulfuric acid in the acid rain sample based on the titration?
-The calculated molarity of sulfuric acid in the acid rain sample was approximately 3.4 x 10^-3 M.
Outlines
🧪 Introduction to Acid-Base Titration
The paragraph introduces a titration experiment involving a 20 mL sample of acid rain and 1.7 mL of 0.0811 M sodium hydroxide (NaOH) to determine the concentration of sulfuric acid (H₂SO₄) in the sample. It explains that titration involves adding a base to an acid and provides context on NaOH as a strong base and H₂SO₄ as a strong acid.
🔍 Understanding the Reaction and Key Concepts
The paragraph delves into the details of the titration process, discussing the mixing of acid and base to form salt and water. It introduces a useful shortcut formula (M1V1 = M2V2) for titrations, emphasizing the need to account for the number of H⁺ and OH⁻ ions in the acids and bases involved.
🔢 Applying the Titration Formula
This paragraph applies the titration formula to the given data, identifying the number of H⁺ ions in H₂SO₄ and OH⁻ ions in NaOH. It lists the volumes and molarities of the acid and base and sets up the equation to solve for the concentration of sulfuric acid in the rain sample.
📏 Simplifying and Solving the Equation
The paragraph continues by solving the titration equation. It highlights that the volumes can be in any consistent unit (mL or L). The equation is simplified and solved for the concentration of sulfuric acid, yielding a final answer of 3.4 x 10⁻³ M H₂SO₄, considering significant figures.
🌈 Understanding the Endpoint and Conclusion
This final paragraph explains the concept of the endpoint in a titration, where a color change indicates the reaction's completion. It summarizes the steps to find the concentration of sulfuric acid and encourages students to use the provided formula for ease. The video concludes with best wishes for students' future tests and quizzes.
Mindmap
Keywords
💡Titration
💡Molarity
💡Sodium Hydroxide (NaOH)
💡Sulfuric Acid (H2SO4)
💡Endpoint
💡Acid Rain
💡Volume
💡Concentration
💡Strong Acid
💡Neutralization Reaction
💡Formula
Highlights
Titration of a 20 mL sample of acid rain required 1.7 mL of a 0.0811 Molarity sodium hydroxide solution.
The assumption is made that the acidity of the rain is due to the presence of sulfuric acid.
Sulfuric acid (H2SO4) is a strong acid and will dissociate completely.
Sodium hydroxide (NaOH) is a strong base and is used in the titration.
The formula m1V1 = m2V2 is used for acids and bases in titration.
The acid's molarity is multiplied by the number of H+ ions it contains.
The base's molarity is multiplied by the number of OH- ions it contains.
For H2SO4, there are two H+ ions, and for NaOH, there is one OH- ion.
The volume of the acid is 20 mL and the volume of the base is 1.7 mL.
The molarity of the base (NaOH) is 0.0811.
The endpoint in a titration is when a color change occurs, signifying the reaction is complete.
The goal is to find the molarity of sulfuric acid in the rain sample.
The formula simplifies the calculation without needing to balance the chemical equation.
The volumes used in the formula do not need to be in liters; they can be in mL.
The calculated molarity of H2SO4 is 3.4 x 10^-3 M.
The importance of using the same units for volumes in the formula is emphasized.
The final answer is presented with an explanation of significant figures.
Transcripts
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