Test 1 #5

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HannahBui 2K
Posts: 101
Joined: Wed Sep 18, 2019 12:19 am

Test 1 #5

Postby HannahBui 2K » Sat Feb 01, 2020 1:45 pm

Hi! Can someone please explain how they solved this question? I got a little confused after setting up the ICE Table. The question states, "The pKb of chloroacetate is 11.2. A 0.5M solution of chloroacetic acid was made for a synthetic procedure that required a weak organic acid. What is the poH of this solution at equilibrium?"

Haley Dveirin 1E
Posts: 101
Joined: Sat Jul 20, 2019 12:17 am

Re: Test 1 #5

Postby Haley Dveirin 1E » Sat Feb 01, 2020 1:56 pm

You can use the pKb to find the Kb and then use the Kb to find the Ka. Then create your ICE table and set that equal to the Ka you found to solve for x which equals [H30+]. The use [H30+] to find pH and use pH to find pOH.

preyasikumar_2L
Posts: 101
Joined: Fri Aug 09, 2019 12:17 am

Re: Test 1 #5

Postby preyasikumar_2L » Sat Feb 01, 2020 3:07 pm

From the pKb you can find the value of Kb. Kb = 10^(-pKb)
From the Kb you can find the value of Ka. Ka = Kw/Kb
You can find the equation of Ka using the reaction and ICE table, and solve for the missing x values of the substances at equilibrium using the value of Ka and the equation.
From there, you'll know what the value of [H3O+] is at equilibrium.
From the [H3O+], you can find the pH. pH = -log[H3O+]
From the pH, you can find the pOH. pOH = 14 - pH.

DesireBrown1J
Posts: 98
Joined: Wed Sep 18, 2019 12:18 am

Re: Test 1 #5

Postby DesireBrown1J » Wed Feb 12, 2020 12:49 pm

My final answer was pOH=12.5. Is that correct (for those who got full points)?


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