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Calculate the pH of a solution prepared by dissolving 0.150 mol of acetic acid and 0.300 mol of sodium acetate in water sufficient to yield 1.00 L of solution. The Ka of acetic acid is 1.76 ⋅ 10-5. Calculate the pH of a solution prepared by dissolving 0.150 mol of acetic acid and 0.300 mol of sodium acetate in water sufficient to yield 1.00 L of solution. The Ka of acetic acid is 1.76 10-5. 3.892 10.158 5.056 2.516 4.502

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Answer:

pH = 5.056

Explanation:

Acetic acid (CH₃COOH), is in equilibrium with water, thus:

CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺

From here, it is possible to obtain H-H equation to find pH of a solution:

pH = pka + log₁₀ [CH₃COO⁻] / [CH₃COOH]

Where molarity of acetic acid, [CH₃COOH], is 0.150mol / 1.00L = 0.150M and molarity of acetate ion, [CH₃COO⁻], is 0.300mol / 1.00L = 0.300M. pKa of the buffer is -log Ka: 4.754

Replacing:

pH = 4.754 + log₁₀ [0.300M] / [0.150M]

pH = 5.056

The study of chemicals and bonds is called chemistry.

The correct answer is 5.056

Acetic acid (CH₃COOH), in the question, is in equilibrium with water, thus:

[tex]CH_3COOH + H_2O <--> CH_3COO^- + H_3O^+[/tex]

From here, we have to find the pH of a solution:

[tex]pH = pka + log_1_0 \frac{ [CH_3COO^-] }{ [CH₃COOH]}[/tex]

All the data is given in the question, put the value and solve it

  • Molarity of acetic acid, [CH₃COOH], is 0.150mol / 1.00L = 0.150M
  • Molarity of acetate ion, [CH₃COO⁻], is 0.300mol / 1.00L = 0.300M.
  • pKa of the buffer is -log Ka: 4.754

Replacing the value:-

[tex]pH = 4.754 + log_1_0 \frac{[0.300M] }{ [0.150M]}[/tex]

After solving the pH is 5.056.

Hence, the correct answer is 5.056.

For more information, refer to the link:-

https://brainly.com/question/13899929

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