A .1M solution of chloroacetic acid is 11% ionized. Using this information, calculate [ClH2COO-], {H+], [ClCH2COOH] and Ka for cloroacetic acid.
To solve this problem, we need to set up an ICE table and use the initial ionization percentage to determine the concentrations of the various species.
Let's define the variables:
[ClH2COO-] = concentration of chloroacetate ion
[H+] = concentration of hydrogen ion (proton)
[ClCH2COOH] = concentration of chloroacetic acid
Ka = acid dissociation constant for chloroacetic acid
First, we need to determine the initial concentration of chloroacetic acid in the solution. A 0.1 M solution means that the original concentration of ClCH2COOH is 0.1 M.
Next, we can set up the ICE table:
ClCH2COOH(aq) ⇌ ClH2COO-(aq) + H+(aq)
I 0.1 M 0 0
C -x +x +x
E 0.1 M - x x x
Based on the given information, we know that 11% of chloroacetic acid is ionized. This means that x (the concentration of ClH2COO- and H+) is 0.11 times the initial concentration of ClCH2COOH, or 0.11 * 0.1 M = 0.011 M.
Therefore:
[ClH2COO-] = 0.011 M
[H+] = 0.011 M
[ClCH2COOH] = 0.1 M - x = 0.1 M - 0.011 M = 0.089 M
Finally, to calculate Ka, we can use the formula for Ka:
Ka = [ClH2COO-] * [H+] / [ClCH2COOH]
Plugging in the values we calculated:
Ka = (0.011 M) * (0.011 M) / (0.089 M)
Ka = 0.001121 / 0.089
Ka = 0.0126 (rounded to four decimal places)
Therefore, the values are:
[ClH2COO-] = 0.011 M
[H+] = 0.011 M
[ClCH2COOH] = 0.089 M
Ka = 0.0126 (rounded to four decimal places)
To solve this problem, we need to understand the relationship between ionization percentage and concentration in a solution.
1. Calculate [ClH2COO-]:
The ionization percentage of 11% means that 11% of the chloroacetic acid has dissociated into ClH2COO-. This also indicates that the remaining 89% is still in the form of ClCH2COOH.
Since we have a 0.1 M solution of chloroacetic acid, we can calculate the concentration of ClH2COO- as follows:
[ClH2COO-] = 0.1 M × 11% = 0.01 M
So, [ClH2COO-] (concentration of chloroacetate ions) is 0.01 M.
2. Calculate [H+]:
Since chloroacetic acid is a weak acid, it does not dissociate completely. However, we can assume that every molecule of ClH2COO- formed will produce one H+ ion. So, the concentration of H+ will be the same as the concentration of ClH2COO-, which is 0.01 M.
So, [H+] (concentration of hydrogen ions) is 0.01 M.
3. Calculate [ClCH2COOH]:
The remaining 89% of the chloroacetic acid is still present in its original form, ClCH2COOH. To calculate its concentration, we subtract the concentration of ClH2COO- from the initial concentration of chloroacetic acid.
[ClCH2COOH] = 0.1 M - 0.01 M = 0.09 M
So, [ClCH2COOH] (concentration of chloroacetic acid) is 0.09 M.
4. Calculate Ka for chloroacetic acid:
The relationship between the concentrations of [H+], [ClH2COO-], and [ClCH2COOH] can be described by the dissociation constant, Ka.
Ka = ([H+][ClH2COO-]) / [ClCH2COOH]
Substituting the values we calculated:
Ka = (0.01 M × 0.01 M) / 0.09 M
Simplifying further:
Ka = 0.0001 M^2 / 0.09 M
Final result:
Ka = 0.0011 M
So, Ka (dissociation constant for chloroacetic acid) is 0.0011 M.
I can't type all of this on one line; lets call chloroacetic acid just HAc(since the Cl plays no role anyway---except of course it makes it a much stronger acid than straight acetic acid).
11% ionized means 0.1M is 0.1 x 0.11 = 0.011M
...........HAc ==> H^+ + Ac^-
.........HAc ==> H^+ + Ac^-
I........0.1M.....0......0
C......-0.011...0.011..0.011
E.....0.1-0.011..0.011..0.011
Substitute those numbers into Ka expression and solve for Ka. I would evaluate equilibrium HAc first. You can insert Cl where ever you need it.