Berapakah harga pH larutan campuran 100 mL larutan 0,6 M CH_(3) COOH dengan 100 mL larutan ...

CoLearn | Bimbel Online 4 SD - 12 SMA
2 Feb 202403:40

Summary

TLDRThis video explains how to calculate the pH of a buffer solution formed by mixing acetic acid (CH3COOH) with barium hydroxide (Ba(OH)2). It walks through the step-by-step process of determining the limiting reagent, performing the chemical reaction, and identifying the leftover weak acid and its conjugate base, forming a buffer. The formula for buffer solutions is then applied to calculate the concentration of H+ ions and, subsequently, the pH. The final pH is found using the -log of the H+ concentration, resulting in a value close to 6.

Takeaways

  • 🧪 The problem asks for the pH of a solution mixing 100 ml of 0.6 M CH3COOH with 100 ml of 0.2 M Ba(OH)2, with Ka of CH3COOH given as 1.6 * 10^-5.
  • 📏 Moles of CH3COOH are calculated as 60 mmol (0.6 M * 100 ml).
  • 📏 Moles of Ba(OH)2 are calculated as 20 mmol (0.2 M * 100 ml).
  • ⚖️ The reaction is 2 CH3COOH + Ba(OH)2 → Ba(CH3COO)2 + 2 H2O.
  • 🔍 Ba(OH)2 is identified as the limiting reagent because 20 mmol / 1 (Ba(OH)2) < 60 mmol / 2 (CH3COOH).
  • 🔬 Ba(OH)2 reacts completely, leaving 20 mmol CH3COOH unreacted and forming 20 mmol Ba(CH3COO)2.
  • 🧫 A buffer solution is formed consisting of the weak acid (CH3COOH) and its conjugate base (CH3COO- from Ba(CH3COO)2).
  • 📊 The concentration of H+ ions is calculated using the buffer equation: [H+] = Ka * [weak acid] / [conjugate base].
  • 🧮 [H+] = (1.6 * 10^-5) * (20 mmol) / (40 mmol) = 8 * 10^-6 M.
  • 📉 The pH of the solution is calculated as pH = -log[H+] = 6 - log 8, giving a pH close to 6.

Q & A

  • What is the first step in calculating the pH of the mixture in this problem?

    -The first step is to determine the moles of CH3COOH and Ba(OH)2 by multiplying their molarity by the volume.

  • How many moles of CH3COOH are present in the mixture?

    -There are 60 mmol of CH3COOH, calculated by multiplying 0.6 M by 100 mL.

  • How many moles of Ba(OH)2 are present in the mixture?

    -There are 20 mmol of Ba(OH)2, calculated by multiplying 0.2 M by 100 mL.

  • What is the limiting reagent in the reaction, and how is it determined?

    -Ba(OH)2 is the limiting reagent, determined by comparing the moles divided by the stoichiometric coefficient (Ba(OH)2 gives 20/1 = 20, CH3COOH gives 60/2 = 30).

  • How much CH3COOH reacts in the mixture?

    -A total of 40 mmol of CH3COOH reacts with Ba(OH)2.

  • How many mmol of CH3COOH remain after the reaction?

    -There are 20 mmol of CH3COOH remaining after the reaction.

  • What is formed after the reaction between CH3COOH and Ba(OH)2?

    -The reaction forms Ba(CH3COO)2 (20 mmol) and water (40 mmol).

  • Why is the resulting solution considered a buffer solution?

    -The solution is considered a buffer because it contains a weak acid (CH3COOH) and its conjugate base (CH3COO-), formed from the salt.

  • How do you calculate the concentration of H+ in this buffer solution?

    -The concentration of H+ is calculated using the formula: [H+] = Ka * (mol of weak acid) / (mol of conjugate base).

  • What is the pH of the buffer solution, and how is it calculated?

    -The pH is calculated as -log[H+], which results in a pH of approximately 6 - log(8).

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pH calculationbuffer solutionacid-base reactionchemistry tutorialacetic acidbarium hydroxidemolarityreaction equationschemistry educationKa value
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