Atmospheric CO2 levels have risen by nearly 20% over the past 40 years from 320 ppm to 400 ppm. (a) Given that the average pH of clean, unpolluted rain today is 5.4, determine the pH of unpolluted rain 40 years ago. Assume that carbonic acid 1H2CO32 formed by the reaction of CO2 and water is the only factor influencing pH. CO21g2 + H2O1l2 Δ H2CO31aq2

Video transcript

Hi everyone, this problem reads production of hydrogen sulfide from sewage waste is projected to increase 19% in the next 20 years from 270 parts per million to 320 parts per million if the current ph of sewage waste due to hydrogen sulfide is 6.78, what will be the ph of sewage sewage waste 20 years from now, assume that only hydrogen sulfide is contributing to the P. H. Okay, so our goal here is to determine what the ph of the sewage is going to be 20 years from now and in this problem we are dealing with gas and so one of the gas laws were going to need to solve this problem is Henry's law, and Henry's law is written as C is equal to K times P where C is our concentration of a dissolved gas, K is Henry's law constant, and P is partial pressure of the gas. So, we're going to want to determine what is K in this equation. And in order for us to solve for K, we first need to calculate, see the concentration of a dissolved gas and p the partial pressure of the gas. So let's go ahead and do that. So, to determine the concentration, we're going to write out our equation. Okay, so we have hydrogen sulfide gas is at equilibrium and it dissociates into its ions. Alright, so the first thing we want to do is solve for the concentration of hydrogen ions. Okay, so let's go ahead and do that, we know that the concentration of hydrogen ions is going to equal 10 to the negative P. H. And then the problem we're told what the ph is. So we can go ahead and plug that in. So we have 10 to the negative 6.78. So when we do this calculation we get the hydro ni um ion concentration is equal to 1.659. Six times 10 to the -7. Okay, so what this tells us is or what we know is that the concentration of hydro knee um ion is equal to the concentration of our conjugate base. Okay. And so this number that we just saw for is going to be the same for both of them. All right. So we know our concentration for hydro knee um ion. We know for our conjugate base. Now we need to solve for the concentration of hydrogen sulfide and we can do that by writing out our K. A. Expression are acid dissociation expression. Okay, so this is going to be K. A. One. So it's going to be our products over our reactant since. And remember our goal here is to solve for our concentration of hydrogen sulfide and we know what our K. is equal to because um it's a value that we can look up. So that value is 8.9 times 10 to the negative eight. So now that we know what our concentration of hydro ni um ion and conjugate bases. We can plug that into our equilibrium expression R. K. Expression to solve for our concentration of hydrogen sulfide. So we're going to go ahead and plug in so because we know our hydro knee um and the conjugate base is the same. The number is going to be squared. And for concentration of hydrogen sulfide it's going to be X minus that number. So X. Is going to actually be our concentration of hydrogen sulfide. So that's what we're going to solve for here. Alright, so let's cross multiply and when we cross multiply we're going to get 8.9 times 10 to the negative eight times X minus 1.6596 times to the negative seven is equal to 1.6596 times 10 to the negative seven squared. Okay, so we just cross multiply. So now we need to foil this out so we can simplify. So once we foil out and simplify we get 8.9 times 10 to the - X -1.5 times 10 to the -14 is equal to 2. times 10 to the -14. So we're gonna go ahead and add 1.5 times to the negative 14 to both sides. So we can get rid of that. So once we simplify this out we get 8.9 times 10 to the negative eight is equal to or this is X 8.9 times 10 to the negative eight X. Is equal to 4.25 times 10 to the negative 14. So, remember we're solving for X. So let's go ahead and divide both sides by 8.9 times 10 to the negative eight. Okay. And when we do that, we get X is equal to 4.78 times 10 to the negative seven which we said is our concentration of hydrogen sulfide. So going back up remember we're trying to solve our Henry's law gas equation. So we needed our concentration. So now we need our partial pressure. Okay, We need our partial pressure of the gas and we know that We have parts per million. So, what this equates to is 270 most of hydrogen sulfide per one times 10 to the sixth moles of waste. Okay, once we do this calculation, we get 0.000270 more percent of hydrogen sulfide. Okay? So for our property of gasses more percent is equal to pressure percent. And so that means our pressure of hydrogen sulfide is going to equal 0. A. T. M. All right. So like we said up above, we know that K is equal to C over P and we have both C and P. So, let's solve for K. So the concentration we solve for is 4.78 times 10. We rearranged this equation. The K. Over the K. Over the K equal C over P. We rearranged this equation up at the top right here. Okay, so we rearranged it to isolate it for K. And that's what we're looking for the gas. The we're looking for Henry's law constant. Alright, So now we're plugging in the concentration and partial pressures we just solved for Okay, so K is equal to 1.76 times 10 to the negative three. Okay, so now that we know that We can calculate the concentration of hydrogen sulfide 20 years from now. Okay, so let's go ahead and make some room. Okay, So we're going to just one second. Alright, So we're going to continue this here. All right, Alright, so now we're going to calculate the concentration 20 years from now, and we have 320 moles of our hydrogen sulfide over one times 10 to the 6th moles of waste. And this is equal to 0.3 to zero more percent of hydrogen sulfide. And remember our property of gasses more percent is equal to two pressure percent. So, our pressure of hydrogen sulfide is equal to 0.3 to 0 M. So, what this means is our concentration of hydrogen sulfide is going to equal R K. So are 1.76 times 10 to the negative three times our pressure. So 0.3 to 0 M. Which gives us 5. times 10 to the negative seven. So that's our concentration of hydrogen sulfide. So now we need to calculate our concentration of hydro ni um 20 years from now because the concentration of hydro knee um is how we're going to get our P. H. Remember P. H. Is equal to negative log of hydro nian concentration. Okay so our concentration of hydro ni um 20 years from now, we're gonna write out our K. Expression. Alright so we have our K. A. Is equal to our products are concentration of products over our concentration of reactant since and we know that this is equal to 8.9 times 10 to the -8. So plugging in our values were going to get X squared over 5.6347 times 10 to the negative seven minus X is equal to 8. times 10 to the negative eight. Ok so this minus X. In the denominator. We can check to see if X is negligible and we can assume that it is because our concentration of hydrogen sulfide is low. So we're going to use the quadratic equation to solve for X. And when we do the quadratic equation to solve for X will get X is equal to 1.8382 times to the negative seven. And like we said P H. Is equal to negative log of hydro ni um concentration and this X. Represents our hydro knee um concentration. So when we plug that value in negative log of 1.8382 times 10 to the negative seven, we're going to get a P. H. Equal to 6.74. And this is our final answer. This is our ph Of the sewage 20 years from now. That's the end of this problem. I hope this was helpful.