Calculate the [H3O+] and pH of each polyprotic acid solution. b. 0.350 M H2C2O4

Video transcript

Hey everyone, welcome back. We're asked to consider a solution of 0.591 molar of carbonic acid and were asked to calculate the concentration of hydrogen ions and the ph of our solution first. If we go ahead and look up our K. S. For our Carbonic acid, we find that our K. A. one Is going to come up to 4.3 Times 10 to the -7. While our K. A. two comes up to 5. Times 10 to the -11. And as we can see when we compare the two K. S. RKA two is extremely small compared to R. K. A. one. So the ph of our solution will only depend on the hydro ni um ions produced from our first equilibrium reaction. So let's go ahead and write that reaction and create our ice chart. We have our carbonic acid and we react this with water and we produce our bicarbonate ion and our hydro ni um ions creating our ice chart. We were told that we started with 0.591 Moeller of our carbonic 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 in our carbonic acid is going to be a minus X. Since we're losing reactant and a plus X in our product side. Since we're gaining products at equilibrium, we have 0.591 minus X. In our react inside and an X. And an X in our product side. Now let's go ahead and use our K. A. To determine the value for X. So we know ourka. was equivalent to 4.3 Times 10 to the -7. And this is going to be equal to our products over our reactant. Now let's go ahead and check if we can disregard our X. In our denominator, we can do so by taking Our 0.591 and dividing this By our K. A. of 4.3 times 10 to the -7. And if we get a value greater than 500 then we can go ahead and disregard our X. And in fact we do get a number greater than 500, we get a value of 1.37 Times 10 to the 5th, Which is greater than 500. So we can go ahead and disregard that X. In our denominator, simplifying this a bit further, we end up with X squared equals 4.3 times 10 to negative seven Times 0. Taking the square root of both sides. We end up with an X. Of 5.04 Times 10 to the negative for Mueller, which is going to be the concentration of our hydro ni um ions. Now let's go ahead and calculate our ph and we can do so by taking the negative log of our concentration of 5.04 Times 10 to the -4. When we do this, we end up with a ph of 3.30. So our final answers here are going to be a ph of 3.30 and a concentration of hydro ni um ions of 5.4 times center negative four Mueller. So I hope this made sense and let us know if you have any questions.