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What is the pH of a 0.1 M solution of acetic acid (CH3COOH) given its Ka is 1.8
What is the pH of a 0.1 M solution of acetic acid (CH3COOH) given its Ka is 1.8 x 10^-5?
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Q1
What is the pH of a 0.1 M solution of acetic acid (CH3COOH) given its Ka is 1.8 x 10^-5?
2.87
4.76
3.87
5.00
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Using the formula for weak acids, pH = 0.5(pKa - log[C]) = 0.5(4.74 - log(0.1)) = 4.76.
Questions & Step-by-step Solutions
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Q
Q: What is the pH of a 0.1 M solution of acetic acid (CH3COOH) given its Ka is 1.8 x 10^-5?
Solution:
Using the formula for weak acids, pH = 0.5(pKa - log[C]) = 0.5(4.74 - log(0.1)) = 4.76.
Steps: 7
Show Steps
Step 1: Identify the concentration of acetic acid (CH3COOH), which is given as 0.1 M.
Step 2: Find the acid dissociation constant (Ka) for acetic acid, which is given as 1.8 x 10^-5.
Step 3: Calculate the pKa using the formula pKa = -log(Ka). For Ka = 1.8 x 10^-5, pKa = -log(1.8 x 10^-5) ≈ 4.74.
Step 4: Use the formula for weak acids to find the pH: pH = 0.5(pKa - log[C]). Here, [C] is the concentration of the acid, which is 0.1 M.
Step 5: Substitute the values into the formula: pH = 0.5(4.74 - log(0.1)).
Step 6: Calculate log(0.1), which is -1. So, pH = 0.5(4.74 - (-1)) = 0.5(4.74 + 1) = 0.5(5.74).
Step 7: Finally, calculate 0.5(5.74) = 2.87. Therefore, the pH of the solution is approximately 4.76.
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