Calculate the pH of a .0550 molar solution of sulfuric acid. Here we're given the K1 and the K2 for sulfuric acid. Now notice that K1 is a pretty large number, a number much greater than one. That's because the first proton of sulfuric acid is of a strong diprotic acid. K2, though, is a number smaller than one. That's because removing the second H+ ion is much more difficult. It's a weaker version of the original strong diprotic acid.
With this information, let's get to it. TE one we're going to use the Bronsted-Lowry definition to predict the product's formed when H2SO4 reacts with water. So here we have H2SO4 aqueous plus H2O liquid and here we're going to say that it's a strong asset. So it reacts completely with the water donating an entire H+ ion to the water molecule to create HSO4- aqueous plus H3O+ aqueous. Now the initial concentration here of the sulfuric acid is .0550 molar. We're going to say sulfuric acid completely dissociates initially, meaning the concentration of sulfuric acid equals the concentration of hydronium ion and the intermediate form. So these would also have the same initial concentration.
With this information we can go to step 2. IN step two we set up an ICE chart for the intermediate form created in step one and reacted also with water. So the intermediate form formed in step one was hydrogen sulfate, so HSO4-. This is also called bisulfate. So here it reacts with water as an acid. Donating an H+ to the water molecule it becomes the sulfate ion and water gaining in H+ becomes the hydronium ion. Here we have is the initial concentration .0550, but also for the hydronium ion. Water is a liquid. Liquids and solids are ignored in an ICE chart. We don't have any information on the basic form.
Now here, using the initial row, we're going to lace the amounts from step one for the intermediate form and the H3O+, which we did, we place a zero for the basic form. We now for step four, we're going to say we lose reactants to make products and we're going to say here that using the change row you're going to place A -, X for the reactants and A + X for the products. With this information, we just bring down everything that we have and that'll give us our equilibrium run. OK, so bringing down everything, this would be oh .0550 -, X + X + X. Now our ICE chart is filled in. We can say using the equilibrium row, set up the equilibrium constant expression with Ka2 and solve for X.
Now we're using Ka2 because now we're talking about losing this last and final H+ of the diprotic species to create our basic form. At the end here we're going to say Ka2 equals products over reactants. So sulfate ion times hydronium ion divided by the bisulfate ion Ka2. When we plug that number in, Ka2 here would equal 1.2 * 10-2. At equilibrium, both of these products are X, so that's X2 divided by 0.0. And actually, it wouldn't just be actually oh .0550 + X, so that would actually be oh .0550 + X * X / .0550 - X.
All right. So as we can see, this is going to be a little bit tricky in terms of solving this expression. We're going to have to set up a quadratic formula in order to isolate X here. Now here. Once we do that, we're going to solve for pH by adding the hydronium ion concentrations that we found in parts one and five to determine the complete H+ concentration or total H+ concentration. All right. So let's get to it. We're going to cross multiply these two together. So basically this distributes here and this distributes here. So when we do that, what are we going to get initially? When we do that, we're going to get initially 6.6 * 10-4 -, 1.2 * 10-2 X equals this X going to distribute here and distribute here. So make sure that's going to give me 0.055. Oh X + X2.
We have all these X variables, but if this X that has the highest exponent of two, so it's our lead term, that means everything has to be moved over to the right side. We're going to add this to both sides here so it adds together with this X, and then we're going to subtract this from both sides. So when we do that, we're going to get X2 + .067 X minus 6.6 * 10-4. This will be my a, my B and my C and with this we can set up the quadratic formula. So still not done. So remember the quadratic formula itself is negative b ± sqrt b2 -, 4 AC over 2A. Let's start lugging in the values that we have, right? So here we said that B is going to be negative 0.067 plus or minus .0672 - 4 * 1. Don't forget the negative sign of C, so -6.6 * 10-4 / 2 * 1.
We're going to solve for everything in here and then take the square root of it. So when we do that, we're going to get 0.067 plus or minus oh .08443 / 2. Because this is plus or minus, that means there's two possible values for X1. Value is if we added .08443 which would give me initially .008715 or we could do minus 0.08443 which would give me negative 075715. Remember the correct X is the one that where you can place it anywhere on the equilibrium row and it gives you back a positive value. The -1 will not work because if you place it here it will give you a negative concentration at equilibrium which is not possible. That means that we cannot utilize this one.
So now we're going to say what is our total H+ concentration. So remember H+ is the same thing as H3O+. It would be equal to this. This is the amount that we had initially from the first setup, step one. And then we just found out what X is. The additional H3O+ that was created O. We're going to say .0550 + X. So that's .0550 plus the X that we found of .008715. When we add those together, we're going to get 0.063715. That is my total amount of H3O+. With that we can find pH which is negative log of H3O+. So plug that in and with this information we can finally find our pH, which comes out to be 1.20. So this would be the final pH of my sulfuric acid solution.