Problem: Calculate the pH titration of 50 mL of 0.02M H3PO4 from pH 1 to 13 with 0.4M NaOH.
So I believe the balanced reaction is:
H3PO4 + OH --> H2PO4 + H2O (Since Na I believe is insignificant, maybe that's wrong. If so then, H3PO4 + NaOH --> NaH2PO4 + H2O)
ICE table:
•Meaning that I is: H3PO4=0.02M, OH=0.4M, H2PO4=0M & H2O is blank (since again insignificant)
•Meaning that C is: H3PO4 (Limit Reagent)=0.02M-0.02M=0M, OH=0.4M-0.02M=0.38, H2PO4=0M+0.02M=0.02M, H2O is blank again
•Meaning that E is: ????????? (Help)
I think I'd used the ICE table correctly and balanced the reaction correctly (if not please tell me). Also I don't know how much volume of OH or H2PO4 I should use to get the E in the ICE table (maybe I'm going about this problem completely wrong I don't know). Then there's how to get the pH titration for pH 1 to 13 with the Henderson Hasselbach formula. All I know at this point is that I H3PO4 has 3 pKa's that are as follows: 2.1, 7.2, and 12.3.
Would you check your problem please? Usually titration curves are done with pH vs volume. One then chooses volumes to make the curve.
I do know that titration curves are done with pH vs Volume. This is pretty much exactly how my teacher worded it, so I do know that I'm reaching that's why I'm trying to figure it out to the best of my ability. Please someone help.
I see you still think changing names helps.
Here is what I think. I believe you are to calculate the pH of various solutions from pH 1 to pH 13 but I don't understand the phrase "pH titration". And I don't know how the volume fits in. I can show you how to do the HH equation from pH 1 and up.
pH = pKa1 + log (b)/(a)
You have 50 mL x 0.02M = 1.00 millimol.
........H3PO4 + OH^- ==> H2PO4^- + H2O
I........1.00....0........0
add..............x..............
C.......1.00-x..-x.........x
E.......1.00-x....0.........x
1 = 2.1 + log (b/a)
1 = 2.1 + log (x/1-x)
x = 0.0736 if I didn't err.
Two points about this.
That is the way you might use the HH equation but I have no idea what that number means. You can calculate the mL of the 0.4M NaOH to get that pH but I don't know where that is supposed to take you.
Point two is that with an H3PO4 solution this dilute any calculation of pH MUST include the use of the quadratic formula (which the HH equation doesn't do) and I wouldn't put much faith in the calculated pH this way anyway. As an example you can substitute 2 for pH and you get almost the same thing for x as when pH = 1. Does that sound right?
I don't know that this will help but I didn't attend the profs lecture and you did. Perhaps I've said something here that will trigger the right response in your mind. The main problem I'm having with this is "what do we do with the number we get from the calculation?" In effect, I'm inclined to ask after the calculation, so what? Good luck.
I appreciate the help even with the slight attitude. Just to let you know in all respect my professor assigned this during my winter break last minute to me soon to be going back for spring semester, we've never learned how to do buffer equations yet. So I'm trying my best here, as for why I don't ask my professor for help isn't the issue, the issue is no matter how many times I email him to request help he refuses to answer my emails (and yes I check the email. So again appreciate the help even with the attitude. Plus, I'm new to this website so I didn't realize that the question stayed visible for quite sometime (reason for the name changing), so thanks.
With sincere respect,
C.
#1. We are here to help.
#2. I have no idea why you think there is an attitude problem. I'm sorry I couldn't answer your question exactly but as I stated I simply don't understand the question completely. I hoped that whatever I posted would trigger a response from your personal experience with the prof.
#3. Everyone here tries to help as completely and as quickly as possible. We try (and hope) that students keep the same screen name because that helps us provide more complete answers if we realize the student is the same or the student is different.
#4. The bottom line is that I couldn't help but remind you that changing screen names wasn't the fastest way to get help.
#5. You might be surprised at the number of poorly worded questions we get.
#6. Come back as often as you need help. We're always open.
And yes, the questions are there for all to see almost forever I suppose. The last time i looked there were pages and pages of questions dating back several years.
This is a confusing question that is poorly worded. So, I am not sure what your professor wants you to do. The confusing part, to me, is However, this is what I think you are suppose to do: determine the volume of 0.4M NaOH added to reach the equivalence point for H3PO4 and determine the pH at each equivalence point. If this is the case then, repost your question and someone will try to help you.
This is a confusing question that is poorly worded. So, I am not sure what your professor wants you to do. The confusing part, to me, is from pH 1 to 13 with 0.4M NaOH. However, this is what I think you are suppose to do: determine the volume of 0.4M NaOH added to reach the equivalence point for H3PO4 and determine the pH at each equivalence point. If this is the case then, repost your question and someone will try to help you.
I don't want to waste my time doing a lot of typing if this is not what you need.
To calculate the pH titration of H3PO4 with NaOH, you are on the right track with the balanced equation:
H3PO4 + OH- → H2PO4- + H2O
Next, let's determine the initial concentrations (E in the ICE table) for each species at pH 1.
Since you have 50 mL of 0.02 M H3PO4, the initial concentration of H3PO4 is 0.02 M. The initial concentration of NaOH (OH-) is 0.4 M, as given in the problem statement.
Now, let's consider the specific steps for calculating the pH at each point from pH 1 to 13:
1. Calculate the amount of NaOH required to reach the desired pH by using the stoichiometry of the balanced equation. Since H3PO4 is tribasic (has three acidic hydrogens), it undergoes multiple titrations.
2. Determine which acidic hydrogen of H3PO4 will be deprotonated at each pH value based on the pKa values (2.1, 7.2, and 12.3). For example, at pH 1, all three hydrogens of H3PO4 are present as H3PO4 since the pH is below all the pKa values.
3. Use the Henderson-Hasselbalch equation to calculate the pH for each step. The Henderson-Hasselbalch equation for a weak acid is:
pH = pKa + log([A-] / [HA])
Where:
- pH represents the desired pH value for the titration step
- pKa represents the acid dissociation constant for the specific acidic hydrogen being deprotonated
- [A-] represents the concentration of the deprotonated form of the acid (in this case, the concentration of H2PO4-)
- [HA] represents the concentration of the protonated form of the acid (in this case, the concentration of H3PO4)
4. Repeat steps 1-3 until you reach pH 13, considering the successive deprotonation of H3PO4 as the pH increases.
By following these steps, you should be able to calculate the pH titration curve for H3PO4 using 0.4 M NaOH from pH 1 to 13.