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: (e) 61.0 mL.

Accepted Answer

Hello everyone today. We have the following question. A 25 millimeter sample of 250.8 Moeller of tri methyl amine with the KB expression for KB value of 6.4 times 10 to the negative fifth is tight traded with 100.40 Mueller of hydro bromide solution. What is the ph after 51 millimeters of the tight trend has been added. So first you want to write our chemical equation. So we know that we have our try methyl amine any acquis form reacting with our hydro Tomic acid or HBR. And this is going to form our tri methyl amine ion and a bromide ion all Aquarius. So the second step is to calculate the initial number of malls. So first we want the moles for our tri methyl amine. So the moles for our tri methyl amine. This is going to be our 25 middle leaders. We are going to convert from milliliters to liters by using the conversion factor that one millimeter is equal to 10 to the negative third leaders. And then we're going to use the conversion factor, we're gonna multiply by the polarity and that is for everyone. Leader, there is .8 moles. And so when our units cancel out, we are left with 2.0 times 10 to the negative 3rd Mueller. And so next we need to find the initial moles for our HBR which we take our middle leaders converted to leaders by using the conversion factor that one leader 0 to 10 to the negative third leaders. And then as before we multiply by the polarity which says that we have 100. moles Per one leader once our units cancel, we're left with 2.04 times 10 to the negative third moles. And third we can set up our I. C. F. Table. So we have our chemical reaction, we have our tri methyl amine reacting with our hbr or hydroponic acid forming us our ion and our bromide ion. So we have our I. R. C. And our F. So our initial concentration for tri methyl amine as we just calculated is going to be too 0.2 times 10 to the negative third moles for hbr It's going to be 2.04 times 10 to the negative third moles. And of course we have zero initial concentrations for our ions that we created. And so when we combine these two we're going to subtract from our initial. So this is going to be minus two times 10 to the negative third. This is gonna be minus two point oh four times 10 to the negative third moles were going to be adding it to the respect of ions. So this would be plus two times 10 to the negative third moles. This is going to be plus 2.04 times 10 to the negative third moles. of course that gives us zero for our first column for a second column. We're actually going to get zero as well. We are going to get four 0.0 times 10 to the negative fifth moles, 1st, 2nd and 3rd, we're going to get 10, 2 times 10 to the negative third moles And 2.04 times 10 to the negative 3rd moles, respectively. And so it's important to know a few key things. The first is that H p r is a strong acid, it completely dissociates. And then we have our tri methyl amine ion, which is a weak acid, so only partially dissociates. And then we have our bromide ion which is for all intents and purposes for this question, a neutral ion. And so therefore the strongest asset will determine the ph. So we're going to use HBR to determine the ph And so what we need to do next is we need to take further concentration of hbr hydrochloric acid. We need to take the amount of moles and divide by our total volume To calculate for our total volume, we're going to take our 25 middle leaders Plus our 51 million years. And this is going to be 76 ml but we have to convert this to leaders are 76. Middle leaders here would just be multiplied by the conversion factor that one. Middle leader is 10 to the negative third leaders are units cancel and we're left with 0.076 leaders. And so now we can calculate for our total volume down here or a concentration of H P R, which is going to be our moles or four point oh times 10 to the negative fifth moles over the volume. We just calculated for 0.76 liters. And that is going to give us 5.263, 2 times 10 to the -4 moller. Now. And lastly we have to calculate the ph ph of our strong acid is going to be negative log of our hbr hydroponic acid. And so we're simply going to plug in negative log and our concentration that we sold for 5.26 32 times 10 to the negative four moller. And we plug this into our calculator. We receive a final ph of 3.29 as our answer. 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: (e) 61.0 mL.

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