Consider the reaction:
HC2H3O2(aq) + H2O(l) ⇌ H3O+(aq) + C2H3O2-(...

Question

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?

Accepted Answer

Hi everyone for this problem, it says the following equilibrium reaction has an equilibrium constant of 6.5 times 10 to the -5. 298 Kelvin calculate the equilibrium concentration of H 30 plus at 298 kelvin when the solution contains 2980.5 to 5 moller of hc seven H 502 at the beginning. Okay, so our mission here is to calculate the equilibrium concentration of H 30 plus. So because we're dealing with equilibrium here, we're going to want to create an ice table in order to solve for our equilibrium concentration. So let's rewrite out our reaction. Okay, so we're going to I'm going to draw this line here to separate our reactant from our products and we're going to write out ice on the side. So remember that pure solids and pure liquids are not included in the equilibrium expression. So we're going to ignore water here and we're told that Our initial concentration of RHC 7 H is 0.5-5. So that goes in our Iroh, I represents our initial concentration. So that means that our initial concentration is going to be zero for these. And since that's the case, this means our reaction in order for it to reach equilibrium, it's going to want to shift to the right because it has no products and it's going to want to shift towards the side with note that is zero. So that equilibrium is reached. What that means for our change in concentration is our reactant is going to decrease and our products is going to increase. So because we have one mole of everything, we have minus X. For our change in concentration for our reactant and then plus X for both of our products. Okay, R E represents our equilibrium concentrations, so we're just going to add up both of our first two rows and bring it down, so 0.5 to five minus X. And then this is just X. So our ice table is complete. In order for us to solve for our equilibrium concentration, we're going to use this equilibrium row of our table but we need to solve for X in order for us to know what our value is for our equilibrium for each of our products and react ints. So what we can do to solve for X is use our equilibrium expression which is K. C. Is equal to the concentration of our products over the concentration of our reactant in our problem, we're told what our K. C value is, it's 6.5 times 10 to the negative five. So let's go ahead and plug that in here. So our K C is equal to 6.5 times 10 to the negative five. And this is set equal to our concentration of products over reactant. So we're going to take what's in our equilibrium row and plug it in here. So for our products we have H 30 plus and c seven H minus and those are both represented by X. So we have our products is X times X raised to their stoke E. O metric coefficient. So we only have one mole of each. So this is raised to the it just has a coefficient of one. Our reactant is H C seven H 502. And our equilibrium row says that it is 0.5 to five minus X. Okay, so our goal here, like I said, is to solve for X. And one thing that we can do right now is check to see if this X and the denominator is negligible. If it is negligible we can go ahead and ignore that X and remove it. But if it's not it needs to stay there. The way that we check if it's negligible is by dividing our initial concentration and our K. C. So our initial concentration is 0. to five and our K C is 6.5 times 10 to the negative five. So right here is X negligible and I'm not sure if I'm spelling that right, so negligible. Okay, So this is how we check if this value is greater than 500, then X is negligible, but if it's not then it means X is not negligible. So this value does end up being greater than 500. So that means that this x right here is negligible. Okay, so we can eliminate it and let's go ahead and rewrite out our K. C expression. So we have 6.5 times 10 to the negative five is equal to X times X which is X squared over 0. to five. So because X was negligible, we got rid of that X. But if it was not it would stay there. So now let's go ahead and simplify this and solve for X squared. We can multiply both sides of our equation by 0.5 to five. So we'll get X squared is equal to 6.5 times to the negative five times 0.5 to five. Okay, so when we simplify that we get 3. five times 10 to the negative five we're going to take the square root of both sides to get rid of the X squared. And so we get a final answer of X is equal to 5.8 417 times 10 to the -3. So the problem is asking us for our equilibrium concentration of H plus. So take a look at our ice table, we see that the equilibrium concentration of H 30 plus is represented by X. So here I'm going to highlight that. So that means whatever our X value is, that's what our equilibrium concentration for H 30 plus is going to be. And so that means it is just going to be the value of X. That we just saw for. So our equilibrium concentration of H. 30 plus is equal to X. Okay, and our X we solved four is 5. times 10 to the negative three molar. This is going to be our final answer. That is the equilibrium concentration of H30 plus. That's the end of this problem. I hope this was helpful.

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?

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