Calculate the pH and the percent dissociation of the hydrated
ca...

Question

Calculate the pH and the percent dissociation of the hydrated 
cation in 0.020 M solutions of the following substances. 
See Appendix C for values of equilibrium constants. 
(a) Fe1NO322

Accepted Answer

hi everyone for this problem. It reads determine the ph and percent ionization of a 0.65 molar solution of chromium three. The P. K. A. Of hydrated chromium three is 3.95. So the question that we want to answer here is the P. H. And the percent ionization. So let's go ahead and start off by creating an ice table. Because this is what's going to help us to calculate the ph. Okay, so to create our ice table, we're first going to write a reaction. So we're going to write a reaction of the ion, the hex aqua chromium three ion reacting with water. So what that looks like is the following. Okay so this is our reaction and we're going to create an ice table. Okay? And for an ice table remember that? We don't include solids or liquids. So we can go ahead and ignore the water. That's right there. So we know what our initial concentration is. The eye in our eyes table stands for initial concentration and we know that the initial concentration is 0.065 moller. Okay. And we have no product. So that means our reaction is moving in the forward direction and that means our concentration of reactant is going to decrease and our concentration of products is going to increase. So for our change which is the c row of our ice table for reactant, we're going to have a minus because our concentration is decreasing and this is going to be X. Because we only have one mole of it for our products. Our concentration is increasing. So this is going to be positive and we only have one mole of this product. So it's just X. And then the same thing for the second product which is our hydro ni um plus X. So when we combine both rose we get 0. minus X. And then we just bring the two X is down. So now we can figure out the K. A. Okay, so the K. A. Wasn't given but the P K. Was given in this problem and we can find the K. A. From the P. K. So K. A. Is equal to 10 to the negative P. K. A. And we know what the P. K. A. Is. We're told it is 3.95. So using that information let's go ahead and solve for the K. So R. K. Is going to equal 10 to the negative 3.95. And when we solve for this, the value we get is 1.122 times 10 to the negative four. So this is our K. A. Value. So now let's go ahead and write R. K. Expression, our equilibrium constant expression. And it is going to equal the concentration of products over the concentration of reactant. And we're gonna take what we have in our ice table too, replace those with the equilibrium. Okay, so when we rewrite this out, our concentration of products is our two products and our reactant. Remember we don't include solids or liquids and either our rice table or our equilibrium constant expression. So we're just going to have one reactant here. Okay. And so what we're gonna do is we're going to replace what we have in our ice table, the equilibrium row of our ice table into this problem, we already know what R. K. Is R K. A. Is we just saw for it. It's 1.122 times 10 to the negative four. And now we're going to replace what's in the equilibrium row of our ice table with those values. Okay, so what we're going to get is X times X over 0.65 minus X. So our goal here is to solve for X. And when we look at our ice table, our value from X is our value for the hydro knee um ion concentration. Once we know what the value is, we can then solve for P. H. So let's go ahead and move forward with solving for X. Okay, so let's first check to see if the X. And this denominator here is negligible. It's going to be negligible. If the value of the initial concentration which is 0.65 divided by the K. A. Value 1.122 times 10 to the negative four. It's negligible. If this value is greater than 500. Okay, so this value is greater than 500. So this X is negligible and we can ignore it. So when we rewrite this what this now becomes is 1.122 times 10 to the negative four is equal to X squared over 0.65. Okay so now what we're going to do is we're going to solve for X. Okay and when we solve for X we're going to get X is equal to 2.7006 times 10 to the negative three. And this is equal to our hydro knee. Um ion concentration. So now that we know that value we can solve for P. H. Because the equation for ph is ph is equal to the negative blog of hydro knee um ion concentration. So now we just need to plug the value of X. N. Because the value of X is our hydro ni um ion concentration. So what we get then is P. H. Is equal to negative log of 2.7006 times 10 to the negative three. And when we saw for this we get P. H. Is equal to 2.57. So that is answer the first part or the first answer of what we need to answer. The second thing that we need to answer is present ionization. Okay so percent ionization is going to equal are concentration at equivalent at equilibrium for this. So this is the equilibrium concentration over the initial concentration. Okay so that is going to be our product over our reactant. Our concentration of product over our concentration of reacting. And we're going to multiply that by 100. So let's go ahead and plug these values in. So we're going to get percent. Ionization is equal to 2.7006 times 10 to the negative three molar, divided by 0. molar Times 100. So our final answer for percent ionization Is equal to 4.2%. So this is the second answer for this problem and that is it for this problem. I hope this was helpful.

Calculate the pH and the percent dissociation of the hydrated cation in 0.020 M solutions of the following substances. See Appendix C for values of equilibrium constants. (a) Fe1NO322

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