Hi, I would like to ask you that if we want to find the concentration of phosphoric acid, which volume at equivalent point should be used. And I wonder that why did you find the concentration of H3PO4 by using the volume of NaOH at the first equivalent point?
I believe what you had as Ka2 is actually your Pka2 unless I am missing something. This video helped me quite a bit but I still am trying to figure out my calculations for the second part of the curve
In the original solution that we are titrating there is going to be some H2PO4- as well as HPO4-2 and some PO4-3 (that will quickly react to form HPO4-2). The bottom line here in this titration is that we are adding OH- to titrate the excess protons or H3O+ that were generated from the weak acid H3PO4. Whether we are actually titrating the H3O+ in solution due to some of the weak acid, H3PO4, dissociating with water or the H3PO4 itself it does not matter in our calculations because all of the acidic components of the initial weak acid solution COME from the weak acid. So in these calculations we can think that we are starting with 100% undissociated H3PO4 to keep things simplified BUT we are really titrating all the acid components of the solution that comes from the weak acid and thus we can predict all points on on a titration curve because 1 OH- reacts with 1 H+. Either way the amount of OH- added will drive the reaction stoichiometrically no matter if we "think" that we are starting with 100% undissociated H3PO4 or not.
Hi, I can't seem to figure out how to find the second concentration value, which would be the concentration of HPO4(2-). The video does not go over the second equivalence point concentration for the acid, or does it? Maybe I am missing it
+julian garcia There is no second concentration value. The concentration of the acid (which is based on 40 ml of acid that I used) is the concentration of the acid. I found the concentration of the acid to be 0.034. This means that at the first equivalence point 0.034 moles of protons are removed from 0.034 moles of acid per 1 liter of solution. At the second equivalence point you will remove another 0.034 moles of protons per liter and so on. If you used the second equivalence point to obtain the concentration of the acid you would need twice the volume of the base and you would remove twice the amount of moles of protons (from the 1st and 2nd equivalence points) and would get the same answer.
I don't have the graphic, but have the Ka1, 2 and 3, how can I calculate the 3 Ph that I need? The first one and the last one, I understood from your lecture, but the second ph (4.8)
There are several ways and the easiest way to plot if you have the Ka's would be the following: 1st equivalence : 1/2 (pKa1 + pKa2) 2nd equivalence: 1/2 (pka2 + pKa3)
+Felipe González If you do not have a graph of pH and volume then you will need the concentration of the acid with the concentration of the strong base and the initial volume acid. With those values you can calculate the moles of OH- needed to neutralize the acid because H+ and OH- will be equal at each equivalence point. The difference here is there is three hydrogens to be pulled of by the strong base, therefore you need 3 times the OH- volume to reach complete neutralization. If you have 100 ml of 1M concentration of phosphoric acid then you need 100 ml of a 1M concentration of strong base (OH-) to reach the first equivalence. You would need another 100 ml of the base to reach the second and so forth. THis works because we are matching the mole of H+ with moles of OH-.
Hi, I'm creating a theoretical titration curve for phosphoric acid with NaOH in excel, and am having trouble calculating the concentrations of H2PO4 and HPO4 in the first buffer region in order to calculate the pH due to a typo in my analytical chemistry book. I have the concentration of both the weak acid and strong base and have a first equivalence point at 6.607 mL NaOH and a second equivalence point at 13.214 mL NaOH but am not sure how to manipulate the information i have to calculate the concentrations in between and need 7 data points in between
Jillyn, your question is what I cover in the lecture. You should be able to calculate each data point for the titration. Have you learned to use the Henderson Hasselbalch equation? This would simplify your work. So by knowing how much hydroxides are added (volume) you should be able to stiochiometrically figure out the amount of conjugate acid remaining and the amount of conjugate base formed. That ratio is put in the Henderson Hasselbalch equation to get the pH for each hydroxide volume point. The derivation of this equation is based on the Ka of the acid and you should know that when the conjugate base and acid concentrations are equal your pH equals the pKa which gives you the Ka of the acid.
If we are talking about molarity which is based on volume (L) of solution then temperature will affect the volumes and thus the molarity of these solutions as they expand or contract due to temperature changes.
Dina Atoum We are not using an indicator. We are using pH values obtained by a probe which are plotted against volume of NaOH (strong base). The graph will indicate the equivalence points or max buffering positions (where pH = pka) by its corresponding shape.
Hello, wouldn't you use N(a)M(a)V(a) = N(b)M(b)V(b) in order to solve for the concentration of the acid H3PO4? N(a) being the number of H+ ions the acid can donate and N(b) being the number of H+ ions the base can accept (# of OH- ions in the base formula) ?
Doing the assignment,20ml of 0.5M sodium acetate buffer with a pH of 4.3 was prepared. A) what is the pH of the buffer if 10ml of 0.1M HCl is added? B) what is the pH of the buffer if 10ml of 0.1M NaOH is added? C) what is the pH of pure water if 10ml of 0.1M HCl is added to 100ml?
There are several ways and the easiest way to plot if you have the Ka's would be the following: You can determine the pKa's graphically by determining the pH at the half equivalence points. 1st equivalence : 1/2 (pKa1 + pKa2) 2nd equivalence: 1/2 (pka2 + pKa3)
No the OH- will pull the H+ from the weak acid. If the acid was in water without the strong base your reaction would have more validity. The hydroxide will drive the reaction that would otherwise not occur very much due to the small Ka of the second reaction. The hydroxide forces the weak acid, H2PO4 - to become HPO4-2 and becomes water in the process. This is how we get a second equivalence point because we neutralized the H2PO4 and the pH at this equivalence is due to the reaction with the conjugate base HPO4-2 with water to make H3O+. Adding a strong base will NOT cause the solution to be acidic as you have suggested.
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Hi, I would like to ask you that if we want to find the concentration of phosphoric acid, which volume at equivalent point should be used. And I wonder that why did you find the concentration of H3PO4 by using the volume of NaOH at the first equivalent point?
I believe what you had as Ka2 is actually your Pka2 unless I am missing something. This video helped me quite a bit but I still am trying to figure out my calculations for the second part of the curve
very nice, How can find TA
: FA Ratio between,if add.Zn Metal Nitrite(No3)
Hi, How do you know that there is no H2PO4- at the start of the reaction? Water is basic to H3PO4 and could pull off the H from the H3PO4.
In the original solution that we are titrating there is going to be some H2PO4- as well as HPO4-2 and some PO4-3 (that will quickly react to form HPO4-2). The bottom line here in this titration is that we are adding OH- to titrate the excess protons or H3O+ that were generated from the weak acid H3PO4. Whether we are actually titrating the H3O+ in solution due to some of the weak acid, H3PO4, dissociating with water or the H3PO4 itself it does not matter in our calculations because all of the acidic components of the initial weak acid solution COME from the weak acid. So in these calculations we can think that we are starting with 100% undissociated H3PO4 to keep things simplified BUT we are really titrating all the acid components of the solution that comes from the weak acid and thus we can predict all points on on a titration curve because 1 OH- reacts with 1 H+. Either way the amount of OH- added will drive the reaction stoichiometrically no matter if we "think" that we are starting with 100% undissociated H3PO4 or not.
Hi, I can't seem to figure out how to find the second concentration value, which would be the concentration of HPO4(2-). The video does not go over the second equivalence point concentration for the acid, or does it? Maybe I am missing it
+julian garcia There is no second concentration value. The concentration of the acid (which is based on 40 ml of acid that I used) is the concentration of the acid. I found the concentration of the acid to be 0.034. This means that at the first equivalence point 0.034 moles of protons are removed from 0.034 moles of acid per 1 liter of solution. At the second equivalence point you will remove another 0.034 moles of protons per liter and so on. If you used the second equivalence point to obtain the concentration of the acid you would need twice the volume of the base and you would remove twice the amount of moles of protons (from the 1st and 2nd equivalence points) and would get the same answer.
I don't have the graphic, but have the Ka1, 2 and 3, how can I calculate the 3 Ph that I need? The first one and the last one, I understood from your lecture, but the second ph (4.8)
There are several ways and the easiest way to plot if you have the Ka's would be the following:
1st equivalence : 1/2 (pKa1 + pKa2)
2nd equivalence: 1/2 (pka2 + pKa3)
Hi. How to find the volume of first and second end point if you don't have a graphic? What calculations you must do to find them?
+Felipe González If you do not have a graph of pH and volume then you will need the concentration of the acid with the concentration of the strong base and the initial volume acid. With those values you can calculate the moles of OH- needed to neutralize the acid because H+ and OH- will be equal at each equivalence point. The difference here is there is three hydrogens to be pulled of by the strong base, therefore you need 3 times the OH- volume to reach complete neutralization.
If you have 100 ml of 1M concentration of phosphoric acid then you need 100 ml of a 1M concentration of strong base (OH-) to reach the first equivalence. You would need another 100 ml of the base to reach the second and so forth. THis works because we are matching the mole of H+ with moles of OH-.
Hi, I'm creating a theoretical titration curve for phosphoric acid with NaOH in excel, and am having trouble calculating the concentrations of H2PO4 and HPO4 in the first buffer region in order to calculate the pH due to a typo in my analytical chemistry book. I have the concentration of both the weak acid and strong base and have a first equivalence point at 6.607 mL NaOH and a second equivalence point at 13.214 mL NaOH but am not sure how to manipulate the information i have to calculate the concentrations in between and need 7 data points in between
Jillyn, your question is what I cover in the lecture. You should be able to calculate each data point for the titration. Have you learned to use the Henderson Hasselbalch equation? This would simplify your work. So by knowing how much hydroxides are added (volume) you should be able to stiochiometrically figure out the amount of conjugate acid remaining and the amount of conjugate base formed. That ratio is put in the Henderson Hasselbalch equation to get the pH for each hydroxide volume point. The derivation of this equation is based on the Ka of the acid and you should know that when the conjugate base and acid concentrations are equal your pH equals the pKa which gives you the Ka of the acid.
what if the volume is in millimolar should i convert it to ml first ?
Thank you so much!
You're Welcome!
how can temperature effect the concentration or the amount of volume needed to titrate with the phosphoric acid?
If we are talking about molarity which is based on volume (L) of solution then temperature will affect the volumes and thus the molarity of these solutions as they expand or contract due to temperature changes.
Where is the indicator that you use and the effect of it on the concentration of h3po4????????
Dina Atoum We are not using an indicator. We are using pH values obtained by a probe which are plotted against volume of NaOH (strong base). The graph will indicate the equivalence points or max buffering positions (where pH = pka) by its corresponding shape.
Hello, wouldn't you use N(a)M(a)V(a) = N(b)M(b)V(b) in order to solve for the concentration of the acid H3PO4? N(a) being the number of H+ ions the acid can donate and N(b) being the number of H+ ions the base can accept (# of OH- ions in the base formula) ?
You used dimensional analysis which is pretty much the same thing but didn't account for N(a)? Is there a reason why
very easy to understand
Doing the assignment,20ml of 0.5M sodium acetate buffer with a pH of 4.3 was prepared. A) what is the pH of
the buffer if 10ml of 0.1M HCl is added? B) what is the pH of the buffer if 10ml of 0.1M
NaOH is added? C) what is the pH of pure water if 10ml of 0.1M HCl is added to 100ml?
How to calculate the pH at the first two equivalence points?
There are several ways and the easiest way to plot if you have the Ka's would be the following:
You can determine the pKa's graphically by determining the pH at the half equivalence points.
1st equivalence : 1/2 (pKa1 + pKa2)
2nd equivalence: 1/2 (pka2 + pKa3)
Hi..why cant we calculate concentration of H3PO4 by using initial ph and Pka1 ? H3PO4-->H2PO4^- + H^+
then [H3PO4] = [H+]^2 ÷ Ka1 ryt?
Did I say that you cannot? Its been a while since I made this. Yes you can and your work is correct.
IN second reaction we should get H3O+ NOT H2O.
H2PO4- + OH- ---> HPO42- + H3O+
No the OH- will pull the H+ from the weak acid. If the acid was in water without the strong base your reaction would have more validity. The hydroxide will drive the reaction that would otherwise not occur very much due to the small Ka of the second reaction. The hydroxide forces the weak acid, H2PO4 - to become HPO4-2 and becomes water in the process. This is how we get a second equivalence point because we neutralized the H2PO4 and the pH at this equivalence is due to the reaction with the conjugate base HPO4-2 with water to make H3O+. Adding a strong base will NOT cause the solution to be acidic as you have suggested.
Sorry, you right. I appreciate your time, and thanks for good video.