Determine the percent ionization of a 0.125 M HCN solution.

Video transcript

Hey everyone, we're asked to determine the% 0.137 moller of formic acid solution. First we have to recognize that formic acid is a weak acid. So we will have a nice chart. Let's go ahead and write out our reaction. We have formic acid and this reacts with water. We then create our forming ion plus our hydro ni um ions. Next creating our ice chart, we were told that we started off with 0.137 moller of formic acid. And we had zero of our products formed initially. We can go ahead and disregard our water since it is a liquid and our change is going to be a minus X. On our react inside since we're losing reactant and a positive X. On our product side. Since we're gaining products at equilibrium on our react inside, we have 0.137 minus X. And an X. And an X. In our product side. Looking up the K. A. of formic acid, we find that it's 1.8 times 10 to the -4. And we can make this equal to our products over our reactant in order to determine our X. But before we do that, let's go ahead and check if we can disregard our X. In the denominator. And we can do so by taking our 0.137 and dividing it by R. K. A. Of 1.8 times 10 to the negative four. If we get a value greater than 500 then we can go ahead and disregard it. And we do get a value greater than 500 and we get so we can go ahead and disregard the X. Simplifying this a bit further. We end up with X squared equals 1.8 times 10 to the negative four times 0.137. Taking the square root of both sides. We end up with an X. That is equivalent to 4.9659 times 10 to the negative 3rd molar. And this is going to be our concentration of hydro ni um ions. Now to calculate our percent ionization, our percent ionization is going to be equivalent to our protons over our concentration of our asset. And we're going to multiply this by 100 since we are talking in percentages. So plugging in our values we get 4.9659 times 10 to the negative third. And we're going to divide this by our concentration of our formic acid which was 0.137. And we are going to multiply this all by Calculating this out. We end up with a percent ionization of 3.6% which is going to be our final answer. So I hope this made sense. And let us know if you have any questions