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A chemist titrates 210.0mL of a 0.2890M dimethylamine CH32NHsolution with 0.2501M HCl solution at 25°C. Calculate the pH atequivalence. The pKb of dimethylamine ...

Question

A chemist titrates 210.0mL of a 0.2890M dimethylamine CH32NHsolution with 0.2501M HCl solution at 25°C. Calculate the pH atequivalence. The pKb of dimethylamine is 3.27. Round your answer to2 decimal places.

A chemist titrates 210.0mL of a 0.2890M dimethylamine CH32NH solution with 0.2501M HCl solution at 25°C. Calculate the pH at equivalence. The pKb of dimethylamine is 3.27. Round your answer to 2 decimal places.



Answers

Calculate the pH of a solution that is 0.00115 $\mathrm{M}$ in $\mathrm{HCl}$ and 0.0100 $\mathrm{M}$ in $\mathrm{HClO}_{2} .$

Problem. City Province it requires have to find the original concentration off Hydrofluoric acid it And we have to find a concentration off Lawry. I am hydro, neo Mayan and Hydrofluoric acid at equilibrium. Okay, and now we see. We don't know the the initial is so concentration off the Hydrofluoric acid McQueen, Opie ish and from BS, we can calculate the high Roni A Mayan. Yeah, we know 20. And after we called it the higher oniy, a Mayan. So we have to put a br vision off the concentration off the Hydrofluoric acid with a so we can put this in the equation. Seaweed. Oh, I see. We got a and let minus as my similar adult. Our brave up the rivers problem. Now I manage this and as here we know already because what, you're going right from the top. Here we go. Five follow one high 10 83. And then we put the question and we started till we stop the deal, the number with the s. And then we find ah a either original right. And from the equation, we have to fight a regional concentration and then we fight Ah f minus. Okay, So we have here. Okay? And now we hot us. I should have here that That should be cool. Let me see. Boy, hold 30 and minus. Nobody will have here. So a row boy. Oh, to a fly in.

Let's look at a buffer that contains 0.45 Moeller formic acid and 0.63 molars permit. We can calculate the um, pH and concentration of hydrogen ions using the Anderson Hustle back equation. And just as a side note, uh, if you have something like sodium for mate, it's going to completely dissociate in water into for me. And so do you mind. So any concentration of sodium for mate, for example, 6.3, is going to imply an equal concentration of, um, for me, you know what's so for the pH? We have the PKK of former gas it so we can just plug that in and and that to the log of the concentration of the country, get base, which is for made in this case, 0.63 molars over the concentration of weak acid before my gas. In this case, so 0.45 Miller Ph is then 3.74 0.146 one. This is the 3.88 61 which one thinking about significant figures is just 3.9, and that's our final pH. Never confuse this page to sell for the concentration of hydro in your minds. Thanks. The pH equals the negative long of the concentration of Honduran reminds. Rearranging this equation we get the concentration equals 10 to the negative. PH. In our case, this is 10 to the negative 3.9 which one considering significant figures is 1.2 times 10 to the negative for Muller.

Okay we've got a buffer but we also have a fictitious acid. So the first thing we're gonna need to do is find a K. A. For our acid. Okay So we're given some information about our x minus mm We know that we have 614 moles And that's in 25 L. So we have 246 moller. So let's go ahead and write the equation for X minus reacting with water. To give us a check X and O H minus. And we'll set up an ice box. Okay? So .246. We don't care about that. 00 minus X plus X. And plus X. So 246 -X. X. And X. Well we have enough information to figure out what X is. So we're told that our solution of X minus has a ph Of 873. So from that we can get our age plus right 10 to the negative. 873. It's 1.86 Times 10 to the -9 More. So we can use that to find the O H minus one times 10 to minus 14 divided by H plus gives us R O H minus and that's our X. Okay, So now we've got the polarity of our X. And this equation is written for a weak base so KB All right. Is H X times O H minus divided by R X minus. So if we start filling in right, we've got X squared Over 246 -X. So we found out X was 5.37 Times 10 to the -6. So we'll square that They were divided by .246 -5.37 times and the -6 although it's not going to change that number at all. So we'll get our KB is 1.2 Times 10 to the -10. Okay? So for buffers each plus is a. K A times are moles of acid divided by our moles of base. So let's go ahead and use KB to find CA right. So that's just KW over KB. So our K. A. is going to come out to be 8.5 Time sounding -5. So now I'm just gonna plug in here. Okay, I've got 8.5, I'm saying the -5 And our moles of acid was .219 moles over a checks Divided by .614 moles of our x minus. So this gives us 3.0. I'm saying the -5 more H plus. So then if we just take minus the log of that, we'll get our P H and r ph is 4.52

So here we're give an example of a weak acid conjugate based solution. So we can use our Henderson Hasselbach equations, that ph equal to the PKK put the loss of concentration of our conjugate base over the weak acid. So we're given information that our initial ph is equivalent to 8.77 And our concentration of our weak acid is equivalent to 0.110 molars. And our concentration of our weak base, Our continent basis equivalent to 0.220 molars. And we can use this information to find the PKK which we're going to need later. Now substituting everything in. Okay? Mhm. We can find that the peka is equivalent to eight points 8.469. Paying attention to significant figures. Only the log digits after the digits after the logs are significant. So in this case we have our PK here. Now we're going to stress the system and were given that the volume volume of the solution Is equivalent to 0.350 leaders. And we're going to stress the system by adding a strong base in terms of hydroxide. So the only compound that can react with a strong base in this case is our weak acid. So h Y would react with hydroxide to yield. In this case it would actually be a complete reaction. Since there is a significant difference in the case between the two that this would yield water plus the conjugate base. And we can see that the concentration of conjugate base will increase while the concentration of the week asked will decrease. So we have to account for those calculations. So the number of moles of our weak acid is equivalent to final would be equivalent to the initial number of moles, which is just concentration times volume according to the concentration formula. Then we're subtracting in this case, since we're using barium hydroxide, There are two moles of hydroxide contributed per mole barium. So the concentrations, the concentration effect is doubled. So we are adding 0.0015 moles. Uh barium hydroxide so that the concentration effect would be doubled. So 0.11 times .35. Then we're subtracting two times 0.0015. And we can see that our final number of moles is 0.0355 moles. Okay. And our mole number of moles of why minus prime is equivalent to our initial. And we're using in addition sign in this case, since there is an increase in concentration times two times 0015 most. Since all of that you consumed age wise, used to make the counterfeit base. So now we can substitute everything in 0.2, two times .35 Plus two times 0.0015. This is equivalent to 0.0800 moles. And in this case we don't have to use concentration. Since when we express this concentrations, the volumes were just cancel out so we can ignore the concentrations in this case and just express it as moles. So ph equals P. K. Was logged of the concentration of conjugate base modified over the concentration of weak acid modified. And let's remember that our PA. 8.469. Now we substitute our values. And yeah. And we find that the PPP becomes eight points 8 to 2. Which makes sense considering we're adding strong base too, a strong base to a solution which would obviously contribute to raise the ph in this case. And that's it.


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