At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

Calculate the pH of a 0.002M acetic acid solution if it is 2.3% ionised at this dilution, ka=1.8×10^-5.

Sagot :

Answer:

3.11 is the pH of the solution

Explanation:

The ionised form of the acetic acid is the acetate ion. To solve this problem we need to solve H-H equation:

pH = pKa + log [A⁻] / [HA]

Where pH is the pH of the buffer

pKa = -log Ka  = 4.74

[A⁻] = Concentration of acetate ion = 0.002M * 2.3% = 4.6x10⁻⁵M

[HA] = Concentration of acetic acid = 0.002M *(100-2.3)% = 0.001954M

Replacing:

pH = 4.74 + log [4.6x10⁻⁵M] / [0.001954M]

pH = 3.11 is the pH of the solution

We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.