Consider the reaction: HC2H3O2(aq) + H2O(l) ⇌ H3O+(aq) + C2H3O2-(aq) Kc = 1.8 * 10-5 at25°C If a solution initially contains 0.210 M HC2H3O2, what is the equilibrium concentration of H3O + at 25 °C?

Video transcript

Hi, everyone. Welcome back. Our next problem says, consider the reaction HC two H 302, aqueous, both H2O liquid and a reversible reaction to form H 30 plus aqueous plus C two H +302 minus aqueous. And then we have that our equilibrium constant case of strip C equals 1.8 times 10 to the negative fifth at 25 °C. If a solution initially contains 0.210 molar HC two H 302, what is the equilibrium concentration of H 30 plus at 25 °C? A 0.104 molar B 0.208 molar C 0.00194 molar or D 0.00388 molar. So we're given our equilibrium constant, we're given one initial concentration and we need to find the equilibrium concentration of one of our products. So we know we can use that equilibrium constant equation to solve for the unknown. However, we've only got one initial concentration provided for us. So we're missing more than one equilibrium constant equilibrium concentrations to solve for our unknown. And when we're missing more than one equilibrium concentration in a pro problem like this. We use the I CE or ice chart technique. So let's make up our chart for this particular equation. So we have our species, we have in our reactants concentration of H C two, 8302, that's an aqueous species. So we will use it. The other reaction is water, which is pure liquid. So it doesn't go in here, then we have the products. So that would be the concentration of H 30 plus and C two H three. I just forgot my brackets, concentration of C two H 302 minus. And then we make a chart with those on the top and then our letters I see and E down the left side standing for initial change and equilibrium. So we have only one initial concentration given to us in the problem explicitly. And that is the concentration of our reactant HC two H 302, which is 0.210 molar. So we fill that in, in the eye column there with the I row. However, when we think about the fact we're starting with and we're just given initial concentration, we can assume that our products are zero. We're starting from that zero point where we haven't generated any products yet. So we can put zero and zero in the I row for our two products. Then let's think about the change that's going to happen. Here. Well, we have no coefficients. We don't need to worry about that. So therefore, our amount of reactant is going to decrease by X. So we'll put minus X and the amount of our products it's going to in increase by X. So plus X and plus X and the C column for the two products. So we have minus X plus X plus X. Now we get to the E row which is equilibrium concentration. So we know that for our reactant here, that's going to be its initial concentration. 0.210 molar minus X and then are two products. Their equilibrium concentration is going to be X and X. And we need to therefore solve for X because the value we're looking for is the equilibrium concentration of H 30 plus. So we look at the H 30 plus column and we see that its equilibrium constant is just X. So we need to solve for X. So we would use our equilibrium constant equation to do that since we know our equilibrium constant constant. So we'd say K subscript C which equals 1.8 times 10 to the negative fifth equals. And then we set up our concentrations of our products at equilibrium. So it's going to be zero plus X multiplied by zero plus X in the numerator. And that will be divided by the concentration of our reactants. So 0.210 minus X. Now no there's no units in here. We recall that when we do these equilibrium constants, they're officially unit less because they're not so much absolute numbers as there are ratios between the concentration and the concentration at standard conditions, which is one. So the units have already canceled each other out even though we don't think of it that way. Usually, now this would get into a complicated uh issue to solve. We'd have to use a quadratic formula except we can use a simplification method. Here. In this case, we're going to compare the ratio of our, our initial concentration to our equilibrium constant. If that ratio is greater than 500 we can ignore the minus X term. So let's look at what that ratio is. So we'd have our initial concentration which is 0.210 oops. And we're going to divide it by our K value which is 1.8 times 10 to the negative fifth. And that's going to equal approximately just rounding that off 11,200 which is greater than 500. Certainly. So that means we can ignore B minus X. So when we go back to our equation that we've set up, we can approximate it as we've got zero plus X multiplied by zero plus X on the top approximate it as X squared divided by 0.210 ignoring the minus X factor in the denominator. So that gives us an equation that's much easier to solve again, X squared divided by 0.210 it has to equal 1.8 times 10 to the negative fifth. So X squared must equal 3.78 times 10 to the minus six. And then taking the square root of both sides, we have X equals 0.001944. And finally, the very last little bitty step, we recall the equilibrium constant is unit less because each of those uh concentrations in it are actually a ratio concentration, but the concentrations themselves have the units of molar. So let's look over at our answer choices. And indeed, we have that choice. C is 0.00194 moer. So again, we use that ice chart to set up our equation. We managed to simplify it using that ratio being greater than 500 allowing us to ignore the minus X factor. And that gave us our answer of choice. C 0.00194 molar for the equilibrium concentration of H 30 plus at 25 degrees. See you in the next video.