ACID/BASE PH question.

[dorjiako]

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What would the PH of the solution be if just enough HCL were added to the solution in the original problem to consume all of the sodium acetate.
(The Answer = PH of 2.87)

The original problem that they are referring to is: What is the PH of the resulting solution if 4g of sodium acetate (CH3COONa) is dissolved in 0.5L of water?(The PKA of acetic acid is 4.74)?

 

[Kimchii]

Find the concentration of the acetate that dissociated. That value is also the concentration of H+ ions needed. Use that value to find the pH. The concentration of sodium acetate that dissociates depends on its pKa.

 

[dorjiako]

Find the concentration of the acetate that dissociated. That value is also the concentration of H+ ions needed. Use that value to find the pH. The concentration of sodium acetate that dissociates depends on its pKa.

Still did not get the right answer with your above explaination. I think we are not really understanding the question. However, if you solved it and got the answer that I mentioned above, I would love to know how you did it. Thanks

 

[indianjatt]

So, you start off with .10M acetate ion.

CH3COO - + H3O+ ----> CH3COOH + H2O
.10M .10 M .10M

So you end up with .10M of acetic acid.
Use the shortcut of sq root Ma Ka to get to the concentration of H3O+, which comes out to sq root (.10M * 10^-4.74) = .0013489.
Then -log(.0013489) = pH= 2.87

 

[dorjiako]

So, you start off with .10M acetate ion.

CH3COO - + H3O+ ----> CH3COOH + H2O
.10M .10 M .10M

So you end up with .10M of acetic acid.
Use the shortcut of sq root Ma Ka to get to the concentration of H3O+, which comes out to sq root (.10M * 10^-4.74) = .0013489.
Then -log(.0013489) = pH= 2.87

Thanks a lot