Calculate the concentrations of H3O+ and SO4
2- in a solution
p...

Question

Calculate the concentrations of H3O+ and SO4 
2- in a solution 
prepared by mixing equal volumes of 0.2 M HCl and 
0.6 M H2SO41Ka2 for H2SO4 is 1.2 * 10-22.

Accepted Answer

Hey everyone. A solution was prepared from equal volumes of 0.1 50 molar hydrochloric acid and 500.500 molar sulfuric acid determined the resulting concentrations of hydro knee in Catalan and our sulfate an ion. If the second acid dissociation constant K. A. To value For sulfuric acid is 1.2 times 10 to the negative second power. Let's begin by writing out how these two acids dissolve in water. We have hydroponic acid which we should recall is one of our memorized strong assets and it's going to dissolve in water which because it's a strong acid, it will undergo full dissociation into its ions where we form its conjugate base, our bromide an ion as well. And actually we don't need parentheses here, we'll use a different color for the conjugate base as well as our second product being our hydro ni um kati on which is the conjugate acid of water. Recognize that hydroponic acid is acting as our bronze did Lowry acid donating a proton to water. Moving on to our second acid sulfuric acid. Recall that this is also one of our memorized strong acids. So it's going to fully dissociate into its ions and water where we have our conjugate base of sulfuric acid which is our hydrogen sulfate an ion. And then we have again the product being hydro liam as the country get asset of water now recognize that sulfuric acid is a die protic acid and so it has to hydrogen two hydrogen protons that are available for donation. So we would undergo a second reaction where we begin with its conjugate base hydrogen sulfate, which also dissolves in water, but because it's not a strong asset, it's a weak acid, it's going to undergo partial dissociation. So we would have an equilibrium and so again, our hydrogen sulfate acts as a bronze said larry acid, donating a proton to water. Justice before we formed the first product being the conjugate base as our sulfate two minus an eye on the conjugate base of hydrogen sulfate. And then we have hydro knee. Um again, as our second product being the conjugate acid of water. So what we need to do is for our final answers find the concentration of our sulfate an ion and hydro knee. Um So we're going to need to make an ice table because this is an equilibrium Recall that in our ice charts, we do not consider liquids for our initial concentration of our hydrogen sulfate because it's sourced from our sulfuric acid and were given the initial concentration of sulfuric acid in the prompt as .500 moller. The conjugate base hydrogen sulfate of sulfuric acid has an initial concentration, which is half of that. So we would take 0.500 moller divided by two and that would give us its initial concentration as 20.250 moller. Now moving on to our product side for sulfate, we don't know what's initial concentration. So we're going to put zero since it hasn't formed yet before hydro ni um we can know that its initial concentration is going to be the sum of half of our concentrations of our strong assets as given in the prompt. And so writing that out, we have a 30 plus concentration equal to the concentration of hbr divided by two plus the concentration of H two, S. 04 divided by two. And plugging in from the prompt. We have 0.1 50 divided by two for hbr plus. Sorry, this is a plus sign, 0.500 divided by two. And this is going to result in a value of 0.75 moller plus 0.2 50 moller. And adding these together we have a sum that gives us 500.3 25 moller as the initial concentration of hydro knee. Um So 250.3 25. And so for the change in our ice chart we would have minus the coefficient for our reacted and then plus X. Based on the coefficient for our products. So carrying everything down to equilibrium, we have 0.2 50 moller minus X for our react inside and then X four sulfate an ion on the product side. And then 40.3 25 moller plus X on our as our second reactant for hydro ni. Um And so we need to figure out if we can neglect our plus X term to make things simpler. And so recall that we would consider taking the initial concentration of hydrogen sulfate. Sorry that the I should be on the outside divided by R. K. A. Two As given in the prompt. And so plugging in our values, we would have 20.2 50 moller as our initial concentration of hydrogen sulfate divided by R. K. A. Two. Given in the prompt as 1.2 times 10 to the negative second power. This will simplify to a value of 20. Which we can see is much less than 500 which is our base of comparison. And because it's less than 500 we would say therefore we cannot neglect x. And so we must form a quadratic equation. And so forming our quadratic equation. We're going to first recall our formula for K. A. Which is equal to the concentration of products divided by the concentration of our reactant. So plugging in what we know we are given our K. In the prompt as 1.2 times 10 to the negative second power equal to the concentration of our products from our I start where we have four hydro knee um 40. 25 molar plus X. And then for sulfate we have just X. So this is multiplied by X. In our denominator. For our reactant side are hydrogen sulfate, we have 0.2 50 minus X. So we need to simplify this to get a equation equal to zero. So beginning we would multiply both sides by .250 -1. Sorry minus X. And then this would cancel it on the right hand side. We would have the product on our left hand side being the result as three times 10 to the negative third power minus 1.2 times 10 to the negative second power X. This is set equal to what's left on the right hand side. As our numerator now isolated as 100.3 25 plus X times X. And so continuing the simplification, we want everything to be on one side. And so we're going to add 1.2 times 10 to the negative second power X. And then subtract three times 10 to the negative third power so that they cancel out on the right hand side. And what we would have left is 100.3 20. Or rather let's begin with our X term. So we would have X squared because we would also do the distribution here and then we would have plus 0.3 25 X plus 1. times 10 to the negative second power X minus three times 10 to the negative third power equal to zero. So for our quadratic equation we can say that A is a value of our coefficient being one for X. V. Is our coefficient as 10.3. And sorry I just realized that we can actually continue to simplify because we have these two like terms here. So let's continue and we'll have X squared plus 20.3 25 X. Sorry .337 X. When we combine our like terms minus three times 10 to the negative third power equal to zero. And so now it's clear that our A value is our coefficient in front of X being one. R B value is 10.337 and then our C value is three is negative three. Since we have that minus term times 10 to the negative third power. Recall that our quadratic equation is X equal to negative B plus or minus the square root of B squared -4 times a times C. But instead of a square root will just raise it to the one half power. This is all in brackets. And then in our denominator we would divide by two times A. So plugging in everything from our variables, what we would have is negative B. Which is negative 0.337 plus or minus. We have B squared so 0.337 squared minus four times one times C. As negative three times 10 to the negative third power raised to the one half power in our numerator. Then in our denominator we are dividing by two times A. Being one. So we would have a result as two different X values. Where our first x value being negative is negative 0.3456 and our second X value being positive is 0.867858. We want to go with the positive X. Value. And so now we can go back to what we have from our ice chart to find the concentration of hydro ni. Um in which at equilibrium according to our chart it's 0. 25 plus X. Which would be 250.3 25 plus 0.867858. Which would result in a value of 0.333679. Which we can round to about 36 fix as 3.34 times 10 to the negative third power moller. So this would be our first answer as the concentration of hydro knee. Um resulting in our solution of sulfuric acid and hydroponic acid. And then for our second concentration we need this concentration of our sulfate an ion which is equal to X. According to our rice table. And our X value we just found is 0.867858. Which we round to about 366 of 8.68 times 10 to the negative third power moller. Which would be our second answer being the concentration of our sulfate, an ion from our solution of hydrochloric acid and sulfuric acid. Our final answers highlighted in yellow correspond to choice A in the multiple choice. I hope everything I reviewed was clear. If you have any questions please leave them down below. And I'll see everyone in the next practice video.

Calculate the concentrations of H3O+ and SO4 2- in a solution prepared by mixing equal volumes of 0.2 M HCl and 0.6 M H2SO41Ka2 for H2SO4 is 1.2 * 10-22.

Watch next