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