Asked by anonymous
Known:
Cyanaocetic acid
Ka=3.55*10^-3
ionization equation: HC3H2NO2<==> H^+ + C3H2NO2^-
Unknown: pH of 0.4M of Cyanaocetic acid
I can't figure out how to calculate this. I only knew how to calculate the pH value of an acid with a known Ka, but not in this case.
Also, I want to know the second ionization equation of succinic acid because I need to find the Ka expression of this.
Can someone please help me out. Thanks in advance for helping out.
Cyanaocetic acid
Ka=3.55*10^-3
ionization equation: HC3H2NO2<==> H^+ + C3H2NO2^-
Unknown: pH of 0.4M of Cyanaocetic acid
I can't figure out how to calculate this. I only knew how to calculate the pH value of an acid with a known Ka, but not in this case.
Also, I want to know the second ionization equation of succinic acid because I need to find the Ka expression of this.
Can someone please help me out. Thanks in advance for helping out.
Answers
Answered by
DrBob222
If you know how to calculate the pH of solution of a known Ka, then why not this one. It's the same thing. For simplicity, let's call this long molecule just HC.
HC ==> H^+ + C^-
Now (H^+)(C^-)/(HC) = Ka.
Do an ICE chart, calculate (H^+) and pH from there.
Succinic acid.
http://en.wikipedia.org/wiki/Succinic_acid
If we call succinic acid H2Su, then
H2Su ==> H^+ + HSu^- and
k1 = ......
Then
HSu^- ==> H^+ + Su^=
an k2 = ,......
I assume you can take it from there.
HC ==> H^+ + C^-
Now (H^+)(C^-)/(HC) = Ka.
Do an ICE chart, calculate (H^+) and pH from there.
Succinic acid.
http://en.wikipedia.org/wiki/Succinic_acid
If we call succinic acid H2Su, then
H2Su ==> H^+ + HSu^- and
k1 = ......
Then
HSu^- ==> H^+ + Su^=
an k2 = ,......
I assume you can take it from there.
Answered by
anonymous
What's an ice chart? I don't think I had learned about that..
Answered by
DrBob222
Show me how you do the pH of an acid with a Ka.
Answered by
anonymous
For example,
Given:
1.000M of Cyanoacetic acid
pH value of 2
HC3H2NO2<==> H^+ + C3H2NO2^-
Unknown: Ka
pH=-LOG[H^+]
[H^+]=10^(-2)
[H^+}=0.01
Ka=(0.01)(0.01)/(1-0.01)
=1.01*10^-4?
Is it right?
Actually I don't know why [H^+] and [C3H2NO2^-] have the same value, also why [HC3H2NO2] equals initial concentration minus [H^+]?
Thanks alot
Given:
1.000M of Cyanoacetic acid
pH value of 2
HC3H2NO2<==> H^+ + C3H2NO2^-
Unknown: Ka
pH=-LOG[H^+]
[H^+]=10^(-2)
[H^+}=0.01
Ka=(0.01)(0.01)/(1-0.01)
=1.01*10^-4?
Is it right?
Actually I don't know why [H^+] and [C3H2NO2^-] have the same value, also why [HC3H2NO2] equals initial concentration minus [H^+]?
Thanks alot
Answered by
anonymous
I think I said something wrong.. I mean I know how to calculate the Ka with a known pH value...sorry..
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