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What is the pH of a solution made from 1 L of 0.05 M acetic acid (CH₃COOH, Ka=1.8x10⁻⁵) mixed with 500 mL of 1 M acetate (CH₃COO⁻)?

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Final answer:

The pH of the solution is approximately 5.756.

Step-by-step explanation:

The pH of the solution can be calculated using the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa of the weak acid and the ratio of the concentrations of the weak acid and its conjugate base.

For this problem, the acetic acid (CH₃COOH) is the weak acid and the acetate (CH₃COO⁻) is the conjugate base. The pKa of acetic acid is given as 4.756.

Using the Henderson-Hasselbalch equation, the pH of the solution can be calculated as:

pH = pKa + log([A-]/[HA])

Plugging in the given values, [A-] = 0.5 M (since 500 mL of 1 M acetate is added) and [HA] = 0.05 M (since 1 L of 0.05 M acetic acid is used):

pH = 4.756 + log(0.5/0.05) = 4.756 + log(10) = 4.756 + 1 = 5.756

Therefore, the pH of the solution will be approximately 5.756.

User Ismail Sahin
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