A 0.100 M solution of chloroacetic acid 1ClCH2COOH2
is 11.0% ion...

Question

A 0.100 M solution of chloroacetic acid 1ClCH2COOH2 
is 11.0% ionized. Using this information, calculate 
3ClCH2COO-4, 3H+4, 3ClCH2COOH4, and Ka for chloroacetic 
acid.

Accepted Answer

Hey everyone, we're asked if a 0.200 Moeller agreed. A solution of Cloris acid is 23% ionized. What is the concentration of Cloris acid? Our hydrogen ion, our chloride ion and K A. For Cloris acid. First, we can go ahead and determine the concentration of our hydrogen ion by using the following formula percent. Ionization is equivalent to the concentration of our hydrogen ion over the concentration of our acid initially times 100% plugging in these values, we have 23.0% equals the concentration of our hydrogen ion divided by 0.200 moller times solving for the concentration of our hydrogen ion. We end up with a value of 0.46 Mohler. Now let's go ahead and determine the concentration of our chloride ion. So our reaction was Cloris acid and this dissociated into our hydrogen ion plus our chloride ion creating our ice chart. Initially we had 0.200 moller of our Cloris acid and we had zero of our products. Initially, our change is going to be a minus X in our react inside and a plus X on our product side at equilibrium, we will have 0.200 minus X at our react inside and an X. And an X in our product side. Since we already found the concentration of our hydrogen ion to be 0.046 moller. This means that our concentration of our chloride ion is also 0.046 Mohler. As we can see from our reaction above Now to calculate Our concentration of our Cloris acid, we're going to take 0.200 and subtract our X. Plugging in Our X. We get 0.200 -0. Molar. This gets us to a concentration of 0.154 moller. And this is our concentration for Cloris acid. Now to solve for R K. A, all we need to do is get our concentration of our hydrogen ion times our concentration of our chloride ion and divide this all by the concentration of our Cloris acid Plugging in these values, we get 0.046 times 0.046 divided by 0.154. And this gets us to a.k. of 0.01374. And these are going to be our final answers now I hope that made sense. And let us know if you have any questions

A 0.100 M solution of chloroacetic acid 1ClCH2COOH2 is 11.0% ionized. Using this information, calculate 3ClCH2COO-4, 3H+4, 3ClCH2COOH4, and Ka for chloroacetic acid.

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