In the case of very weak acids, 3H3O+ 4 from the dissociation of water is significant compared with 3H3O+ 4 from the dissociation of the weak acid. The sugar substitute saccharin 1C7H5NO3S2, for example, is a very weak acid having Ka = 2.1 * 10-12 and a solubility in water of 348 mg/100 mL. Calculate 3H3O+ 4 in a saturated solution of saccharin. (Hint: Equilibrium equations for the dissociation of saccharin and water must be solved simultaneously.)

Video transcript

Welcome back everyone. When very weak acids dissolve in water, the concentration of hydrogen ions H 30 plus from the auto ionization of water is significant compared to the concentration of hydrogen ions produced from the dissociation of the weak acid hydrogen peroxide. H 202 is a very weak acid with A K A value of 1. times 10 to the negative. 12 power calculate the concentration of hydro ni um in a 0. 54 g per 100 mL solution of hydrogen peroxide, you'll have to solve the equilibrium of water and hydrogen peroxide simultaneously were given our solution of peroxide in terms of grams per milliliter. So we need to convert this to polarity of our peroxide H 202. So in order to solve for the polarity of our peroxide solution, We're going to begin with the given mass per volume of 0.454 g of peroxide Per 100 ml of solution in the denominator given in the prompt. And we want to make note of our molecular weight of peroxide which from our periodic table will C is equal to a value of 34.02 g per mole. And so we're going to multiply by this molecular weight as a conversion factor. Where in our denominator, we have 34.02 g of peroxide equivalent to one mole of peroxide in the numerator. From this multiplier, we cannot cancel our units of grams of peroxide in the denominator and grams of peroxide in the numerator. Now we're left with moles per mil leader. But as we stated, polarity is in terms of moles per liter. So we're going to multiply by another conversion factor where we recall that in our numerator, one mL of solution is equivalent to 10 to the negative third power leaders of our solution. And so now we can cancel out milliliters in our numerator with milliliters in our denominator. And we're left with moles of peroxide per liter of solution as our final units. This results in our polarity of our peroxide solution equal to a value of 0.13345 molar. Next, we want to note that we're going to assume that our concentration of hydro ni um which we need to find as our final answer that comes from our peroxide solution. H 202 is equal to X. And we can also assume that our concentration of hydro ni um that comes from water is equal to Y. So we're going to have two different equilibrium is that we need to write out, let's begin with our equilibrium of peroxide and we'll make X the color red and we'll make our Y term the color black. So, beginning with our first equilibrium, we have peroxide, our weak acid which dissolves in water, but it's only going to undergo partial dissociation since the weak acid. So we have an equilibrium where we have our acid which donates a proton to water forming our hydro ni um product and then our conjugate base of peroxide H 02 minus with one less proton. So, plugging in our solved concentration initially for peroxide, we can plug in our value 0. molar for peroxide. So for water, it's a liquid and we will not include this in our ice chart on our product side for Hydro knee. Um as we stated, we're going to begin with the term Y and then for our concentration of our conjugate base of peroxide H 02 minus, we have a initial concentration of zero. Now, as we stated, the concentration of hydro knee um that comes from peroxide is equal to X. So our change is going to be minus X for our reactant peroxide and plus X for our concentration of Hydro Needham's change. And then also plus X for are conjugate base H 02 minus. So now bringing everything down to our equilibrium we have for peroxide 20.13345 minus X. Then we have for hydro ni um X plus Y at equilibrium. And then for H +02 minus, we have just X at equilibrium. So I'm just going to create more room. Now, next to this chart will say that this is step one. Now going to step two, we're going to utilize our given K A for peroxide. We're told that our K A of peroxide is from the prompt 1. times 10 to the negative 12 power for peroxide recall that R K is equal to our concentration of our Aquarius products being H 02 minus times our concentration of hydro ni um Divided by our concentration of reactant. In this case, our reactant being peroxide H- 202. So incorporating what we have from our ice chart will have our K A 1. times 10 to the negative 12 power equal to in our numerator for our concentration of H 02 minus we set at equilibrium. That's X and for a concentration of hydro knee um at equilibrium we have X plus Y. Then in our denominator, our concentration of peroxide we set is 20.13345 minus X. Notice that RKA has a power of 10 to the negative 12 power. This means that our K is small and so we would therefore neglect our minus X term, sorry minus X since it's going to be negligible and sorry just to make this visible. So our minus X term is going to be negligible in comparison to our initial concentration of peroxide. So we can neglect X. So now moving on, we would simplify and have 1.77 times 10 to the negative power equal to X times X plus Y divided by 0.13345 minus X and our denominator. So to simplify further, we're going to multiply both sides and sorry, I accidentally wrote the minus X turn, we're neglecting that. And so now we can multiply both sides by our denominator 0.13345, so that it cancels out on the right. And as a result of our product on the left will have the value 2.3621 times 10 to the negative 13th power set equal to our terms. Now isolated from the numerator X times X plus Y, we're going to call this expression equation one. Now we're going to need to derive a second equation using our third step, which is our ice chart for the equilibrium of water H 20. So we have water acting both as an acid and a base in both of our reactant, which is going to be in equilibrium where we form one mole of hydro knee um as a product and one mole of our H 02 minus an eye on. Now our two reactant of water are liquid form. So in our ice table, we will not be including them, we're going to have initially for hydro ni um X and then for H 02 minus zero, since it's a product then for our change, as we noted above our concentration of hydro knee um that comes from water is going to have a change being why? So the change would be plus Y in this case. So just as we had for our color coordination, we have plus Y here and then plus X here for H 02 minus, we will also have the change being plus Y. And then at equilibrium for hydro ni um we'll have X plus Y and then for H 02 minus will have Y at equilibrium. Now, moving on to step four, we have our KW for water, which we should recall from our textbooks as the value one point oh times 10 to the negative 14th power polarity squared equal to our concentration of hydro ni um multiplied by a concentration of hydroxide. So in terms of our ice table will have one point oh times 10 to the negative 14th power equal to our concentration of our Acquis products. For hydro knee. Um we said that at equilibrium that is X plus Y. And for hydroxide, we said that at equilibrium that it's Y notice that we did not include our reactant because again, they were both liquid water and R K W and K expressions only include our Acquis re agents. So now that we have this association. We will say that this is equation too. and now for step five, we're going to solve for X and Y. And so we're going to have first equation one divided by equation two so that we have and actually I'll use a coal in here. Equation one we stated is on the left hand side are integer 2.3621 times 10 to the negative 13th power divided by equation two integer on the left being one point oh times 10 to the negative 14th power. And we'll just make some more room by moving this over. This is set equal to, for equation one. On the right hand side, we have X times X plus Y divided by our left or right hand side. For equation to being X plus Y times Y notice that in our numerator X plus Y, when we divide is going to cancel out with X plus Y, since it will equal the value one. And so we would simplify the left hand side which will result in a value equal to 23.6 21 equal to what we have left on our right hand side, which would be just X and we don't even need parentheses anymore. So we'll just have X divided by Y left over. So isolating for X, we're going to multiply both sides by Y so that it cancels out on the right, we have X isolated to now equal 23.6, 21 Y. So now we're going to substitute X in equation to is our next step. And as we noted above this is equation to here. So what we'll have is one point oh times 10 to the negative 14th power equal to, let's just make some more room here. Move this over. So that's equal to X which actually we will plug in and substitute as 23.6 to 1 Y plus Y and then we would plug in Y from equation to. So recall that Y plus Y is just going to be Y meaning in our next line, we can simplify so that we have one point oh times 10 to the negative 14th power equal to 23.6 to 1 Y times Y which is now going to simplify when we multiply these two terms we would have on our left hand side still 1.8 times 10 to the negative 14 power. And on our right hand side, 23.62 Y squared and sorry, I'm missing a one. So it's 23.6 to 1 Y squared. Now to isolate for our term, why we're going to take the square root Or rather will begin by dividing both sides by 23.6- and also taking the square root of both sides so that it cancels out our square term on the left and then our 23.6 to 1 will cancel out on the right. And so are Y term isolated will now equal a value of 2.5755 times 10 to the negative eighth power. Sorry, this is times 10 to the negative eighth power. And this represents a concentration. So our unit is still more clarity here. So going back to what we solved for X S since we know that X is equal to 23.6 to 1 Y. And we now know our Y value, we can say that this is equal to 23.6 to 1 times our Y value which we solved above as 2.57, sorry 2.0 5755 times 10 to the negative negative eighth power Moeller. So taking the two products of these terms will find our X value equal to the value 4.8601 times 10 to the negative seventh power moller. And so now we can go back to our concentration of hydro knee um to get our final answer in which from our ice chart in both of our ice charts, it was equal to X plus wide equilibrium. And so we just solved to our value X as 4.8601 times 10 to the negative seventh Power Mueller Plus Why? Which we solved above two, equal 2.5755 times 10 to the negative eighth power Moeller. So taking the sum here, we would get a final answer equal to the value 5.0, sorry five point 06586 times 10 to the negative seventh power moller which we can round 23 sig figs as 5.7 times 10 to the negative seventh power moller. This would be our final answer five point oh seven times 10 to the negative seven power moller as our concentration of hydro ni um in our solution of peroxide. This is going to correspond to choice in the multiple choice. I hope this helped and let us know if you have any questions.