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.
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.
What's an ice chart? I don't think I had learned about that..
Show me how you do the pH of an acid with a Ka.
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
I think I said something wrong.. I mean I know how to calculate the Ka with a known pH value...sorry..
To calculate the pH of a weak acid like cyanocetic acid with a known Ka value, you can use the following steps:
Step 1: Write the ionization equation for the acid:
HC3H2NO2 <==> H+ + C3H2NO2^-
Step 2: Set up an ICE (Initial, Change, Equilibrium) table. Since the initial concentration of the acid is given as 0.4 M, you can assume the initial concentration of the H+ ion and the conjugate base C3H2NO2^- are zero.
HC3H2NO2 H+ C3H2NO2^-
I 0.4 M 0 0
C -x +x +x
E 0.4 - x x x
Step 3: Use the equilibrium expression for Ka to set up an equation:
Ka = [H+][C3H2NO2^-] / [HC3H2NO2]
Step 4: Substitute the equilibrium concentrations into the Ka expression:
Ka = x * x / (0.4 - x)
Step 5: Since the value of x is small compared to 0.4, you can assume that (0.4 - x) ≈ 0.4. This approximation simplifies the equation to:
Ka = x^2 / 0.4
Step 6: Rearrange the equation to solve for x:
x^2 = Ka * 0.4
x = √(Ka * 0.4)
Step 7: Calculate the value of x using the given Ka value:
x = √(3.55*10^-3 * 0.4)
Step 8: Calculate the pH:
pH = -log[H+]
pH = -log(x)
Now, to find the second ionization equation of succinic acid and the Ka expression, follow these steps:
Step 1: Write the ionization equation for the acid. Succinic acid (HOOC-CH2-CH2-COOH) has two ionizable protons. The first ionization occurs as follows:
HOOC-CH2-CH2-COOH <==> H+ + OOC-CH2-CH2-COOH^-
Step 2: The second ionization occurs as follows:
OOC-CH2-CH2-COOH^- <==> H+ + COO-CH2-CH2-COOH^2-
The Ka expression for the first ionization can be written as:
Ka1 = [H+][OOC-CH2-CH2-COOH^-] / [HOOC-CH2-CH2-COOH]
The Ka expression for the second ionization can be written as:
Ka2 = [H+][COO-CH2-CH2-COOH^2-] / [OOC-CH2-CH2-COOH^-]
Note: The concentrations of the species in the above expressions are omitted for simplicity.
I hope this helps! Let me know if you have any further questions.