Here it says calculate the pH of 100 MLS of 250 zero molar carbonic acid with 120 MLS of 0.250 molar sodium hydroxide are added. Now here we're giving K1 and K2 of our carbonic acid solution. What we do first is we need to calculate the equivalence volume needed to by the strong species to reach the first equivalence point. This will tell us how much we need to get to the first equivalence point and all successive ones afterwards.

Now here we're going to do M acid times V acid equals M base times V base. Our strong species is the sodium hydroxide. So here we do of the acid, 250 molar times 100 MLS equals 0.250 molar and volume of the base. We're looking for that. We divide both sides by 0.250 molar and the volume of the base needed to reach the first equivalence point is 100 MPs. This will help us determine how much tightrin was actually reacting in terms of this titration.

So we're going to say volume of strong speech to reacting equals your actual volume. So the volume given to us in the question minus your equivalence volume. So if we did this, it'd be 120 MLS. -100 MLS. There'll be 20 amounts. Now why is that? Well, here, when we calculated the first equivalence volume, this tells me how much volume of NaOH is needed to get to the first equivalence point, which is 100. But in reality, we've gone beyond the first equivalence point because we have a volume greater than 100. We have 120 MLS, so we've gone beyond the first equivalence point.

The 2nd equivalence point would just be another 100 on top of that. So we need 200 MLS to get to the second equivalent point. We're not quite there. So here we're going to use steps one to three to set up the ICF chart. We're going to say before the first equivalence point, volume. You start with the acidic form and PK one after the first equivalence point volume, you're going to start with the intermediate form and PKA two. We are beyond the first equivalence point volume at 120 miles. So we're dealing with the intermediate form of carbonic acid, which is bicarbonate.

So HCL₃ minus here, it's still reacting with base. And over here we get our sodium bicarbonate. Actually we get our this would donate in H so we get water here and then we have NA₂CO₃ here. We don't really worry too much about balancing of this equation. We have this equation here given to us, the sodium's or spectator ions, so they don't really matter. We don't need to actually include them if we desire. We're going to say that this is an ICF chart since it's weak and strong mixing together. So we have initial change. Final remember with an ICF chart we use moles as the units.

So the moles of my intermediate will come from the moles of the original die protic species. So remember moles is liters times molarity. So divide the MLS by 1000 and multiply by the molarity, and we'll have the moles of our intermediate form, so .025 moles. Then we would take the reacting volume of our strong species, which is 20 ML divided by 1000, and then multiply that by the original molarity of the strong base. When we do that, we get 0.005 moles, our conjugate base. Here we don't have anything of it originally, so and then here water the liquid so we ignore it.

Now look at the reactant side. The smaller moles will subtract from the larger moles, so we'd have 0.020 moles left of my intermediate form, 0 left of my strong species. Now based on the law of conservation of mass, whatever we lose as reactants, we gain as products. So this would be plus .005 moles. Now what do we have left? At the end? We have the intermediate form and then we have the final basic form. And here just to simplify, we'll just keep it as CL₃2 minus. So what we fundamentally have here is we have basin acid remaining, so we have a buffer.

So Step 4 says the Henderson Hasselbeck equation is used for a buffer to find the pH of a solution. Using the final row, use the moles of your weak acid and its conjugate base to find the pH. Here we are after the first equivalence point, so we're using PK2. So we're going to say PK2 plus log of base over acid. So here our PKA two is negative log of 5.6 * 10^-11 wizard base, the one with less hydrogen. So that would be the carbonate ion, so 0.005 moles. And then who is our acid form? The one with one extra H plus, so that's our bicarbonate form. So that's 0.020 moles. Plug this into our calculator and we get 9.65 as the pH for this titration.