Calculate the pH of each of the following solutions (Ka and Kb va...

Question

Calculate the pH of each of the following solutions (Ka and Kb values are given in Appendix D): (b) 0.100 M hydrogen chromate ion 1HCrO4-2

Accepted Answer

hi everyone for this problem, it reads what is the P. H. Of a 0.15 molar hydrogen Selena ion? If it's K and KB values are the following respectively. So for this problem we want to solve for ph and recall that P. H. Is equal to the negative log of hydro knee um ion concentration. So riding out our reaction, we have hydrogen selene night ion is going to react with water and at equilibrium we're going to have hydro ni um ion plus the silly night ion. Okay, so our goal here because we know that ph is equal to the negative log of hydro knee um ion concentration we want to solve for the concentration of hydro ni um ion. Okay, so we need to write out our K. A. Expression. Alright and R K A. R acid dissociation expression is equal to the concentration of our products over the concentration of our reactant. And remember for our K expression we do not include liquids. So that high that water is not going to be included. So our concentration of products over our concentration of reactant. Okay. And we actually know what R. K. A. Value is because it was given in the problem we're told it is 4.8 times 10 to the negative nine. And remember our goal here is to solve for our hydro ni um ion concentration. So what we're going to do here is plug in X. For our numerator. Ok, so we're gonna have X times X over because our concentration of hydro ni um ions is going to equal the concentration of our of our Selena I and II on. Okay. So both are represented by X. Which means our reacting are hydrogen Selena ion is going to be the initial concentration which was 0.15 minus X. Okay, so let's go ahead and write that here. So we're going to have 0.15 minus X. All right. And that's going to be set equal to the value of R. K. A. Which we said was 4.8 times 10 to the negative nine. So when we solve for X here because we know X is equal to our concentration of hydrogen ions. When we solve for X, we're going to be able to know what is that concentration. Okay and from there we can solve for P H. So let's solve for X. One thing we need to do is check if X is negligible because we have a minus X here at the denominator. So if X is negligible, what that means is we're going to be able to just cancel out that X. This minus X. And it will just be 10.15 But in order for us to see if it's negligible, we're going to need to first. So let's write that here is X negligible. Okay and the way we do that is by taking our initial concentration which is 0.15 and dividing it by R. K A value which is 4.8 times 10 to the negative nine. If this number is greater than 500 then X is negligible. And when we do the math it is greater than 500. So yes X is negligible. So we can go ahead and ignore this minus X in the numerator. So let's rewrite out our equation. Now with it simplified. So we have X squared over 0. equals 4.8 times 10 to the negative nine. So let's go ahead and solve for X. We'll isolate it and keep it on one side. So X squared is going to equal 0.15 times 4.8 times 10 to the negative nine. Okay, so once we simplify the right side we get X squared is equal to 7.2 times 10 to the negative 10. We want to get rid of the square root on the left side. So we're going to or we want to get rid of the square on the left side. So we're going to take the square root of both sides. So X is going to equal the square root of 7.2 times to the negative 10. So we get X is equal to 2.68 times 10 to the negative five which remember X is also equal to our concentration of hydro knee um ion. So now we have what we need to plug into our equation that we wrote up at the top which is P H. Is equal to the negative log of our hydro knee um concentration. So we get P. H. Is equal to the negative log Of the value we just solved for. So 2.68 times 10 to the -5. Okay, so our final answer is going to be our P. H. Is equal to 4.57. Okay, and this is our final answer. This is the P. H. Of the hydrogen Selena ion. That's it for this problem. I hope this was helpful.

Calculate the pH of each of the following solutions (Ka and Kb values are given in Appendix D): (b) 0.100 M hydrogen chromate ion 1HCrO4-2

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