The question: (a) When the pH of 0.10 M HClO2(aq) was measured, it was found to be 1.2. What are the values of Ka and pKa of chlorous acid?
I got the values as Ka = 0.107 and pKa = 0.97 but the back of the book says these are incorrect.
Did anyone else get these answers? If not could someone explain where I may have made the error. Thank you!
Textbook Question 6D.3
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Re: Textbook Question 6D.3
So we know that the pH of 0.10M HClO is 1.2.
We can write out the reaction as HClO2 + H2O --> H3O + ClO2.
We can also set up our ICE Table:
HClO2 + H2O --> H3O + ClO2.
0.10 M 0 0
-x +x +x
0.10-x x x
This means that Ka = ([H3O][ClO2]) / (HClO2) or (x^2) / (0.10-x)
To find our X value, we can use the antilog of the pH. So our x would equal 10^-pH, in this case, it would be 10^-1.2
After finding our x value, we can plug it into our Ka equation, to get the Ka value.
Once you find Ka, you can find the pKa by using -log(Ka)
I hope this makes sense :)
We can write out the reaction as HClO2 + H2O --> H3O + ClO2.
We can also set up our ICE Table:
HClO2 + H2O --> H3O + ClO2.
0.10 M 0 0
-x +x +x
0.10-x x x
This means that Ka = ([H3O][ClO2]) / (HClO2) or (x^2) / (0.10-x)
To find our X value, we can use the antilog of the pH. So our x would equal 10^-pH, in this case, it would be 10^-1.2
After finding our x value, we can plug it into our Ka equation, to get the Ka value.
Once you find Ka, you can find the pKa by using -log(Ka)
I hope this makes sense :)
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- Posts: 37
- Joined: Mon Jan 09, 2023 10:10 am
Re: Textbook Question 6D.3
Hi,
Could you clarify what Ka equation you used? I found x to be 0.05391 but now I'm stuck. I know there's the equation pKa = pH + log(base/acid). Is that applicable here?
Thanks so much in advance
Could you clarify what Ka equation you used? I found x to be 0.05391 but now I'm stuck. I know there's the equation pKa = pH + log(base/acid). Is that applicable here?
Thanks so much in advance
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