Consider the titration of 30.0 mL of 0.050 M NH3 with
0.025 M HC...

Question

Consider the titration of 30.0 mL of 0.050 M NH3 with 
0.025 M HCl. Calculate the pH after the following volumes 
of titrant have been added: (b) 20.0 mL.

Accepted Answer

Hello everyone today we have the following problem. A 25 mL sample of 0.80 Mueller tri methyl amine with a base dissociation constant of 6.4 times 10 to the negative fifth is tight traded with 100.40 moller of hydra bronek acid solution. What is the ph after 15 ml of the Titan has been added. So the first thing you do is you want to draw out what this chemical reaction would look like. So our chemical reaction would have our tri methyl amine in the acquis form of course reacting with our hydro Tomic acid solution. Also Aquarius forming our tri methyl ammonium ion which is also Aquarius as well as a bromide ion in the acquis form as well. So the next thing we must do is you must calculate the initial moles of our reactant. So first we will find the moles of our try metal. I mean And we're gonna start by using the volume that we were given which was 25 mil leaders. And then we're gonna use the conversion factor that one militia leader is equal to 10 to the negative third leaders. And next we have to multiply by the polarity that we were given which is in terms of most per liter. So we can just say that this is 0.80 malls over one leader. Our units from middle leaders and leaders canceled out giving us two times 10 to the negative third moles of our tri methyl amine. And as for our moles of our hydrodynamic acid, We're going to start with the volume. Once again we added 15 ml as before. We're going to use the conversion factor that one ml equal to 10 to the negative third leaders. And then we're gonna multiply by the morality of the HBR which is going to be .040 moles over one leader. Our units of leaders and middle leaders will cancel out leaving us with six times to the negative fourth moles of hbr. And then we have our two times two times 10 to the negative third moles of our trim ethel. I mean so we're gonna save those numbers for later. Next we have to construct our I. C. F. Table. We're going to have our reaction, our tri methyl amine reacting with our hydroponic acid to form our tri methyl ammonium ion and our bromide ion. We're gonna have our initial concentration are change and our final, So the initial concentrations are going to be our initial moles for trade. Methyl ammonium is going to be two times 10 to the negative third. And then for hydroponic is six times 10 to the negative fourth which we just calculated for of course our initial concentrations for the ions are going to be zero. However we're going to be tight trading with our hydroponic acid. So we're going to be subtracting the number of moles of that hydrochloric acid from both of our reactant and adding it to our products. Our final is just going to be our initial plus our change. So we do that for the first column. We're going to get 1.4 times 10 to the negative third moles. We're going to get zero for hydrochloric acid of course. And then both of our products are going to have six times 10 to the fourth moles. So once we've done that, we must then calculate our total volume that we're going to have. And so we're going to take our 25 ml And add that to our 15 ml. Of course we're going to get 40 ml but we must convert that to leaders. So we're going to get the confusion conversion factor once again that one ml equal to 10 to the negative third leaders. When our units cancel that, we are left with 0.04 leaders as our total volume. So we have our moles and we have our volume and so now we can find our concentration of the ions that we have. It's gonna be a unit of modularity. So our concentration for our tri methyl ammonium is going to be moles over liters. And we calculated that as 1.4 times 10 to the negative third moles over our leaders, which was 0.4 leaders. And this would give us 0.35 molar. And for our tri metal, I mean try metal ammonium, We're going to get six times 10 to the negative 3rd moles Over our 0.04 L. And this is going to give us . small clarity. And so we have those two values. Next. In this multi step process, we are going to find our based association constant or we're going to use that constant for our tri methyl amine, It's 6.4 times 10 to the 5th. And we're going to use that to eventually find our K. A. Value. Which brings us to our next step. We have to find our acid dissociation constant using our KW or art association constant. So we're gonna say K W. Is equal to our K. A. Plus our K. B, solving for K. A. We get K. Is equal to K W over K. B. And we simply plug in our values. Now before we do that however, we must just this herman what concentrations are gonna be plugging in? So for R K. A value, it's going to be the concentration of our trim. Actually mean times are concentration of hydro um ions over the concentration of our tri metal ammonium ion. Using our K. A. Value, We're going to get 1.5, 6 Times 10 to the -10 is equal to 0.35 moller times our concentration of hydrogen ions over our 0. moller giving us the concentration of hydrogen ions As 6.7 times 10 to the -11 moller. And lastly, what we're gonna do with that number is we're gonna calculate for ph we're going to see the ph is equal to our negative log times the concentration of hydrogen ions. And we're simply gonna plug in our value, negative log is equal to 6.7 times 10 to the -11. And we're gonna end up with a final answer of 10.17. And with that, we've answered the question overall, I hope this helped, and until next time.

Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added: (b) 20.0 mL.

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