Test 1 problem 5

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Saachi_Kotia_4E
Posts: 68
Joined: Fri Sep 28, 2018 12:23 am

Test 1 problem 5

Postby Saachi_Kotia_4E » Fri Mar 15, 2019 9:58 pm

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
Posts: 54
Joined: Fri Sep 28, 2018 12:17 am

Re: Test 1 problem 5

Postby keyaluo4C » Fri Mar 15, 2019 10:10 pm

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
Posts: 68
Joined: Fri Sep 28, 2018 12:23 am

Re: Test 1 problem 5

Postby Saachi_Kotia_4E » Fri Mar 15, 2019 10:34 pm

keyaluo4C wrote: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
Posts: 54
Joined: Fri Sep 28, 2018 12:17 am

Re: Test 1 problem 5

Postby keyaluo4C » Fri Mar 15, 2019 11:37 pm

I got pH=8.59

Destiny Diaz 4D
Posts: 51
Joined: Fri Sep 28, 2018 12:28 am

Re: Test 1 problem 5

Postby Destiny Diaz 4D » Sat Mar 16, 2019 4:06 pm

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

Chem_Mod
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Re: Test 1 problem 5

Postby Chem_Mod » Sat Mar 16, 2019 4:13 pm

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
Posts: 31
Joined: Fri Sep 28, 2018 12:20 am

Re: Test 1 problem 5

Postby Helen Zhao 1F » Sat Mar 16, 2019 4:17 pm

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.


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