Calculate the percent ionization of a 0.15 M benzoic acid solutio...

Question

Calculate the percent ionization of a 0.15 M benzoic acid solution 
in pure water and in a solution containing 0.10 M sodium 
benzoate. Why does the percent ionization differ significantly 
in the two solutions?

Accepted Answer

Hey everyone, we're asked, what is the percent ionization of 0.1 12 molar of hydrochloric acid solution in water, what is the percent ionization of 0.12 molar of hydrochloric acid solution in 0.17 molar sodium aside, what is the difference between their percent ionization 1st? Let's go ahead and create our ice chart in pure water. So we know that we have hydrochloric acid and this reacts with water. We then form our azide ion plus our hydro ni um ions creating our ice chart. We know that initially we had 0.112 molar of our hydro's OIC acid and we had zero of our products. Initially we can go ahead and disregard our water since it is a liquid and our change is going to be a minus X. In our react inside and a plus X on our product side at equilibrium we have 0.12 minus X. In our react inside and an X. And an X in our product side. Now, if we look up the K. A. Of hydrochloric acid, we find that it's 1.9 times 10 to the negative five. Now this is going to be equal to our products over our reactant. And we want to solve for X. We can go ahead and check if our X. And our denominator can be neglected in order to do so we can take our 0.12 and divide this by R. K. A. Of 1.9 times 10 to the negative five. And if we get a value greater than 500 then we can go ahead and disregard it. And in this case we do get a value greater than 500. So we can disregard our X. In our denominator, simplifying this a bit further, we have X squared equals 1.9 times 10 to the negative five. And we multiply 0.12 on both sides, taking the square root of this and solving for X, We get an x. 1.50997 times 10 to the -3 Moller. Now to calculate our percent ionization, We can go ahead and take that 1.50997 times 10 to the negative third. And we can divide this by 0.12 and multiply by 100% and this will get us to a percent ionization of 1.26%. Again the formula for our percent ionization is the concentration of our hydro ni um ions over the concentration of our acid multiplied by 100%. Now let's go ahead and calculate this in sodium azide. Again we had our hydro's OIC acid and we reacted this with water. We then got our azide ion plus our hydro ni um ions creating our ice chart. Initially we had 0.12 molar and we had 0. moller of our azide ion. And we had zero of our hydro odium ions, we can disregard our water since it is a liquid and our change is going to be a minus X on our reactant side and a plus X on our product side. Since we are losing reactant and gaining products at equilibrium, we have 0.12 minus X for our react inside 0.17 plus X. For our azide ion and an X for our hydro nia. My on we're going to do the exact same steps as we did before with our K. A. Of 1.9 times 10 to negative five. This is going to be equal to our products over our reactant. So we have X times 0.17 plus X. All over 0.12 minus X. Again we want to check if our X can be negligible And we can do so by taking our 0.12 and dividing it by 1.9 times 10 to the -5. Since this is the same as we did in the previous section, this is over 500. So we can disregard our X in the denominator and our X in our numerator with our azide ion simplifying this a bit further, We get X is equal to 1.9 times 10 to the - times 0.12. All divided by 0.17. This gets us to an ex of 1.34 times 10 to the -5. Now to calculate our percent ionization, We're going to take that X. That we just calculated for which was 1.34 times 10 to the negative five And divide that by our 0.12 And we'll multiply this by 100%. This will get us to a percent ionization of 0.011%. As we can see, the difference in the ionization constants Is due to the presence of one of the products for the 0.12 molar of hydrochloric acid solution in 0.17 Molar sodium aside, which caused less acid to dissociate. And this is why we have differing percent immunizations. So I hope that made sense. And let us know if you have any questions.

Calculate the percent ionization of a 0.15 M benzoic acid solution in pure water and in a solution containing 0.10 M sodium benzoate. Why does the percent ionization differ significantly in the two solutions?

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