When NO2 is bubbled into water, it is completely converted to HNO3 and HNO2: 2 NO21g2 + H2O1l2S HNO31aq2 + HNO21aq2 Calculate the pH and the concentrations of all species present (H3O+ , OH-, HNO2, NO2 -, and NO3 -) in a solution prepared by dissolving 0.0500 mol of NO2 in 1.00 L of water. Ka for HNO2 is 4.5 * 10-4.

Video transcript

Hello. Everyone in this video. We want to go ahead and calculate for the peach and the concentrations of all species which includes all of these. Right over here in a saturated solution at this temperature. This temperature being at 30 degrees Celsius. And assuming that all chlorine reacts with water. So we have the mass of chlorine and we will first find the initial moles. So basically before equilibrium of our H O C. L. And H C. L. So for the molar mass of C. L. Two, let's equal to two times. And looking at the parent table, we have the mass of 35.45 to from each chlorine. And of course we have cl two so two items of chlorine, which is why we're multiplying this by two. So this gives us a value then of 70. grams per moles. So now finding the concentration of H O C. L. Again, some information that were given to us is of course going to be this right over here which looks to me like it's the density of the solution. This is 0.576 g of cl two. For every 100 millions of solution, we're gonna go ahead and multiply this by the molar mass of cl two. So for every one mole of cl two we have 70.9054 grounds and then we'll go ahead and convert Our middle leaders here and two leaders. So for every ml We have one leader and finally doing the multiple ratio for every one mole of H. O. C. L. We have one mole of cl two. So now for the units cancelation we see that the grounds of Cl two will cancel. The middle layers will cancel the molds of Cl two cancel. And this simply leaves us with the moles over latest, which is exactly the units for concentration. So once I put everything into the calculator, we see my my concentration of H. L. C. L. Is equal to 0. molars. Doing the exact same thing here for the concentration of hcl, this is equal to again we're starting off with was given to 0.576 g of cl two over mL of solution. We're gonna go ahead and multiply this conversion factor here into multiplying this with the molar mass to convert the grams into moles. So for every one mole of cl two here we have 70.9054 g. Again we're gonna convert then are milliliters into leaders. So for every 1000 mL we have one leader and finally doing the multiple ratio for every one mole of H. C. L. We have one mole of cl two. So we see that the grams will cancel the male layers will cancel the most of cl two, cancel leaving us with the units of mole of hcl over one leader or leaders. So once you put that into the calculator received that. My concentration for H. C. L. Is equal to 0.81235 molar hcl is a strong asset which associates completely and that reaction is being H. C. L. Reacting with H 20. Which yields H 30. Plus and cl minus. So basically the concentration of hcl is equal to the concentration of H 30 plus, which is also equal to the concentration of cl minus. Since we have a 1 to 1 ratio of each And this equals to 0.081235 molars. Now H oh cl is a weak acid. When you go help go ahead and look up the K. Value for this. So we do. So the K. A. value four H. O. C. L. Is equal to 2.95 times 10 to the negative eight. So let's go ahead and do a dice table. So I. C. E. We have the equation to be H. O. C. L reacting with H 20. Which goes ahead to yield H 30 plus and oh cl minus. So for I this stands for initial see signs for change in the sense for equilibrium. So my initial concentration of H. L cl that's 0.81235 molars for each show this is zero and nothing will happen here for the rest of it because we're dealing with a liquid substance here for H zero plus, this is 0.81235 molar. And for O. C. L minus this is again zero. So for my change here this is unknown for my product side. This will always be a negative or minus. And for my product side this will always be plus. Since we don't know any of these values will have a minus X. Here for my products and then we'll have a plus X. And plus X. For each of my products. Now for my H. Oh cl here for my equilibrium part of this icebox we're just basically combining the initial and they change so for H. L. Cl we have 0.81235 molar minus X. For H. 30. Plus again we're just combining everything. So we have 0.81235 plus X. And then for my O. C. L minus is simply just X. Let's go ahead and scroll down here. So we know that my K. A value is equal to the concentration of H. 30. Plus, multiplied by the concentration of O. C. L minus divided by the concentration of O. H. Or H. Oh cl So we know some of these values. Just go ahead and plug those in. So from R. K. That's equal to 2.95 times 10 to the negative eight. My concentration of H. 30. Plus that's 0. plus X. What's applying this by X. and dividing by let's see. We have 0. -1 terms to basically from this term here and this term here. So we do. So we go ahead and get 2.95 times 10 to the negative eight. Equaling to 0.81235 multiplied by X. And divide by 0.81235. So once we can go ahead and isolate the X. Here Look at that X is equal to 2.95 times to the -8 molars. So the equilibrium concentration of H. L. Cl is then 0.08. Actually let's put, go ahead put this in writing. So the equilibrium concentration of H. O. C. L is equal to 0.81235 minus X. But the value of X again is very small so it won't make a difference whether we subtracted or not. And this is the same for the concentration of H. 30. Plus. So for the concentration of H 30 plus multiplied by the concentration of O h minus this equals to eur kw value. So that's 1.0 times 10 to the negative 14. We'll go ahead and plug in. The concentration of H 30. Plus, so that's 0.81235 multiplied by the concentration of O H minus. Again, this equals to 1.0 times 10 to the negative 14. We want to go ahead and isolate before the concentration of O H minus, which means I'm dividing both sides by 0.81235. If we do so we get the concentration to be equal to 1.23 times 10 to the negative 13. So this means that my final concentrations then, is that the concentration of H. 30 plus is equal to 0. molars. The concentration of O. H minus is equal to 1.23 times 10 to the negative 13 molars. The concentration of H oh cl Is equal to 0.0812 molars. Let's go ahead and scroll down. The concentration of O. C l minus is equal to 2.95 times 10 to the negative eight molars. And the concentration of cl minus is equal to 0.812 molars. And lastly we need to go ahead and solve for P H. So P H. Is equal to the negative log of the concentration of H plus. We have this concentration, luckily. So let's go ahead and plug that in. So we'll take the negative log of 0. molars. Once we do so we get them at Ph is then you go to 1.090. So my final concentrations are all listed here for each of our species, and my ph is equal to 1.090. And this here is all going to be my final answer for this problem.