help with homework 12.57

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905084274
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Joined: Fri Sep 28, 2018 12:22 am

help with homework 12.57

Postby 905084274 » Mon Jan 21, 2019 8:49 pm

can someone explain step by step how you solve 12.57 6th edition. I think the solutions manual is incorrect or is just confusing...

Madeline Ho 1C
Posts: 37
Joined: Mon Apr 23, 2018 3:00 am

Re: help with homework 12.57

Postby Madeline Ho 1C » Tue Jan 22, 2019 12:03 am

For part a, you're given that the pH of 0.10 M HCLO2 is 1.2. pH is -log[H3O+], so to solve for [H3O+] you have to do 10^-pH. In this case, it would be 10^-1.2=0.06 M H3O+. Then, you solve for the Ka (acidity constant), which is (0.06^2)/(0.1-0.06). The pKa is the -log of your Ka value. Part b is the exact same process except [OH-] because C3H7NH2 is a base. Hope that helps!

Mukil_Pari_2I
Posts: 87
Joined: Fri Sep 28, 2018 12:29 am

Re: help with homework 12.57

Postby Mukil_Pari_2I » Tue Jan 22, 2019 12:06 pm

Madeline Ho 1C wrote:For part a, you're given that the pH of 0.10 M HCLO2 is 1.2. pH is -log[H3O+], so to solve for [H3O+] you have to do 10^-pH. In this case, it would be 10^-1.2=0.06 M H3O+. Then, you solve for the Ka (acidity constant), which is (0.06^2)/(0.1-0.06). The pKa is the -log of your Ka value. Part b is the exact same process except [OH-] because C3H7NH2 is a base. Hope that helps!


Do we also only use 1 sf throughout the calculation bc the pH has one sf or do we do take into account the sfs at the end?


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