6B.9

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Jessica Esparza 2H
Posts: 130
Joined: Wed Sep 18, 2019 12:15 am

6B.9

Postby Jessica Esparza 2H » Tue Dec 03, 2019 10:34 am

The question asks to find the [OH], [Ph], and [OH] concentration with the given information that is 1.50M of [H30+]. You can get ph 0.176 and OH from that as 14.17. However, how would you get the OH concentration? I thought you would use the equation that is [M]= 10^-pOH. However, you get an incorrect number. Why would you not use the negative pOH?

Ryan Chang 1C
Posts: 105
Joined: Sat Aug 24, 2019 12:17 am

Re: 6B.9

Postby Ryan Chang 1C » Tue Dec 03, 2019 11:16 am

You are using the correct formula to find the concentration of OH, which is [M]=10^-POH. Plugging in the numbers you get: 10^-14.176 = 6.668 x 10^-15 M, which is the correct answer.

Michelle Le 1J
Posts: 50
Joined: Fri Aug 09, 2019 12:16 am

Re: 6B.9

Postby Michelle Le 1J » Tue Dec 03, 2019 11:18 am

Yeah, your answer is right. You can check because [H3O+] x [OH-] should be equal to 1.0 x 10-14.

AKatukota
Posts: 100
Joined: Thu Jul 25, 2019 12:18 am

Re: 6B.9

Postby AKatukota » Tue Dec 03, 2019 11:22 am

I had a question on this too. In the textbook solutions, it says that the answer for (ii) is 1.50M for [OH-], the pH = 13.824, and the pOH = 0.176?

Jessica Esparza 2H
Posts: 130
Joined: Wed Sep 18, 2019 12:15 am

Re: 6B.9

Postby Jessica Esparza 2H » Tue Dec 03, 2019 6:07 pm

Ryan Chang 3A wrote:You are using the correct formula to find the concentration of OH, which is [M]=10^-POH. Plugging in the numbers you get: 10^-14.176 = 6.668 x 10^-15 M, which is the correct answer.
The textbook says it would be 1.5x10^-14

Edmund Zhi 2B
Posts: 118
Joined: Sat Jul 20, 2019 12:16 am

Re: 6B.9

Postby Edmund Zhi 2B » Wed Dec 04, 2019 12:25 am

I got an answer that is different from the textbook as well and is the same as the ones you all got. I've checked my work multiple times. Maybe there's an error in the solutions?


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