ph problem

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605357751
Posts: 31
Joined: Wed Sep 18, 2019 12:20 am

ph problem

Postby 605357751 » Sat Dec 07, 2019 9:39 pm

I am working on a practice problem that says:

Calculate the pH of a 0.20M solution of HC2H302 with Ka = . 1.8 x 10^-5

After solving for pH, pH = -log [ H+] ---> pH = -log [ 1.8 x 10^-5] I got 4.74 but the answer is 2.72.

Can someone help me find where I went wrong

KSong_1J
Posts: 101
Joined: Thu Jul 11, 2019 12:17 am

Re: ph problem

Postby KSong_1J » Sat Dec 07, 2019 9:54 pm

Ka is equal to [H+][A-]/[HA] and, for this equation, you can rewrite it as Ka = [H+]^2/[HA] because H+ and A- have the same concentration. To find the pH, you need to take the negative log of the H+ ions. Since they give you Ka and you know the concentration of the acid, you can just plug them in and rearrange the equation to solve for [H+]:
(1.8*10^-5) = [H+]^2/0.20M
[H+] = √(1.8*10^-5)(0.20M)
[H+] = .001897
then you can plug this into the pH equation:
-log(.001897)
this will give you 2.72

605357751
Posts: 31
Joined: Wed Sep 18, 2019 12:20 am

Re: ph problem

Postby 605357751 » Sun Dec 08, 2019 5:12 pm

Thankyou this helped a lot!


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