Test 1 #5

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Joseph Saba
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Joined: Thu Jul 11, 2019 12:16 am

Test 1 #5

Postby Joseph Saba » Thu Jan 30, 2020 1:14 pm

Hi, can someone explain how they got the answer on test 1 for #5? For reference, "The pkb of fluoride is 10.8. A .12 M solution of hydrogen fluoride was made for silicon water processing. What is the pOH of this solution at equilibrium? HF + H2O -> F- + H3O+"

Clara Cho 2K
Posts: 103
Joined: Wed Sep 18, 2019 12:18 am

Re: Test 1 #5

Postby Clara Cho 2K » Thu Jan 30, 2020 1:30 pm

You're given pKb so you can find Kb. However, in the reaction given you need Ka because HF is an acid. Solve for Ka by using Ka*Kb=Kw, K =1*10^-14. Ka is now your equilibrium constant.Then use the given initial concentration of hydrogen fluoride, 0.12M, and complete an ice table using the Ka constant. Once you have the concentration of H30+, use the pH formula which is -log(concentration of H30+), to calculate the pH. Then do 14-pH is find pOH.

Joseph Saba
Posts: 154
Joined: Thu Jul 11, 2019 12:16 am

Re: Test 1 #5

Postby Joseph Saba » Fri Jan 31, 2020 11:01 am

Clara Cho 2K wrote:You're given pKb so you can find Kb. However, in the reaction given you need Ka because HF is an acid. Solve for Ka by using Ka*Kb=Kw, K =1*10^-14. Ka is now your equilibrium constant.Then use the given initial concentration of hydrogen fluoride, 0.12M, and complete an ice table using the Ka constant. Once you have the concentration of H30+, use the pH formula which is -log(concentration of H30+), to calculate the pH. Then do 14-pH is find pOH.

What was your answer at the end?

905373636
Posts: 62
Joined: Thu Aug 01, 2019 12:15 am

Re: Test 1 #5

Postby 905373636 » Fri Jan 31, 2020 12:58 pm

Joseph Saba wrote:
Clara Cho 2K wrote:You're given pKb so you can find Kb. However, in the reaction given you need Ka because HF is an acid. Solve for Ka by using Ka*Kb=Kw, K =1*10^-14. Ka is now your equilibrium constant.Then use the given initial concentration of hydrogen fluoride, 0.12M, and complete an ice table using the Ka constant. Once you have the concentration of H30+, use the pH formula which is -log(concentration of H30+), to calculate the pH. Then do 14-pH is find pOH.

What was your answer at the end?


I got 12.15 pH at the end!

Emma Popescu 1L
Posts: 105
Joined: Wed Sep 11, 2019 12:16 am

Re: Test 1 #5

Postby Emma Popescu 1L » Fri Jan 31, 2020 2:46 pm

I found that the pOH was 12.4

Rodrigo2J
Posts: 102
Joined: Sat Jul 20, 2019 12:16 am

Re: Test 1 #5

Postby Rodrigo2J » Fri Jan 31, 2020 4:30 pm

In general, you need to create an ICE table with the information given. Then convert Pkb to Ka and set up an equation to solve for x. Since the 5% rule is not valid in this situation, you need to use the quadratic formula to solve. Once you find x, you can convert [H3O+] to pH. Then just subtract the pH from 14 to find pOH.

Patricia Cardenas
Posts: 103
Joined: Fri Aug 09, 2019 12:17 am

Re: Test 1 #5

Postby Patricia Cardenas » Sun Feb 02, 2020 3:12 pm

does anyone know the average grade for Test #1?

Brian J Cheng 1I
Posts: 115
Joined: Thu Jul 11, 2019 12:15 am

Re: Test 1 #5

Postby Brian J Cheng 1I » Sun Feb 02, 2020 3:21 pm

Average grades aren't posted to us. Ask your TA or Dr. Lavelle for the averages!


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