Test 1 problem 5

[Saachi_Kotia_4E]

I know its a little late but can someone tell me what the answer for number 5 on test 1? I want to make sure I got it right. I had my test on Wednesday and the problem asked about a 0.02 M solution with a pKa of 4.88. Thanks!

[keyaluo4C]

I have a different version of the test but I got the problem right. Could you post the full equation for the problem?

[Saachi_Kotia_4E]

I have a different version of the test but I got the problem right. Could you post the full equation for the problem?

What is the pH of a 0.020 M solution of sodium propionate? the pKa of c3h6o2 is 4.88

C3H5o2-(aq) + H20(l) --> C3H6O2(aq) + OH-(aq)

I know how to do the problem I just want to make sure I got the answer correct

[keyaluo4C]

I got pH=8.59

[Destiny Diaz 4D]

How would you go about this problem I had some trouble understanding it

[Chem_Mod]

You'd do an ICE table with this reaction

C3H5O2-(aq) + H20(l) --> C3H6O2(aq) + OH-(aq)

You'd plug in the initial concentration of C3H5O2- to be 0.02 M and you'd solve for the Kb of C3H5O2- using Ka*Kb = 10^-14. Once you solve for x, x will be the [OH-]. Taking the negative log of [OH-] will give you the pOH. Subtract pOH from 14 to get the pH.

Hope this helps! :)

[Helen Zhao 1F]

I used the pKa to find pKb.
I used pKb to find Kb.
I then made an ice table and used that to find the concentration of OH-.
I used the concentration of OH- to find the pOH.
Finally, I used pOH to convert to pH.