[SweetAsC12H22O11]

I'm having trouble with a new topic covered in class this week, the titration of (polyprotic) mixtures, particularly the following question:

A pool of waste water (10,000 L) contains 0.5M chlorine (from HCl) and 1.0M phosphorus (from phosphoric acid). How many kg of NaOH_(s) must be added to the pool to bring the pH to 7.2 in order for the water to be safely released into a nearby system?

So, HCl is a strong acid,  H₃PO₄ is a weaker acid, which has 3 pKa's: 2.148, 7.20, and 12.15.

H₃PO₄  H⁺+H₂PO₄^- H⁺+ HPO₄^2- H⁺+PO₄^3-

If I understood right in class, since HCl is a strong acid, it will be titrated first (they will be titrated separately).

Now, the desired pH is equal to the second pKa of H₃PO₄. Since pH=pKa at 1/2 Ve, the volume of NaOH added must be equal to 1.5Ve (1.5 x volume at equivalence point). However, I get stuck right in the beginning as I can't figure out how to find Ve without the concentration of the NaOH being used to titrate... Usually it is a simple C1V1=C2V2 calculation...

Can anyone help me figure out what I'm missing? Thanks!

[asmcriminal]

Okay, this is what I did...

Neutralize the HCl

HCl
(0.5moles/1L )*(10,000L) = 5,000 moles HCl

The polyprodic acid has a Pka of 7.2(our target pH). At the second equivalence point.  The hendersonal Hasselbach equation states pH = pka + log([A-]/[HA]).  Where A- is our conjugate base and HA is our acidic proton donor.

H₃PO₄ H₂PO₄^-  pka = 2.148
H₂PO₄^- HPO₄^2-  pka = 7.2

H₂PO₄^- = Conjugate acid
 HPO₄^2- = Conjugate base


So we need the conjugate base A-, and the conjugate acid(HA) to be equal to one, because log (1/1) = log(1) = 0. So we will have PH = pKa + 0 so the pKa and the pH are the same at the half equivalence point. This means we need complete deprotination of the first acidic proton.

So we have H3PO4(1.0M)  1.0 moles/L * 10,000L = 10,000 moles H+ from H3PO4. After we do this, we will have H2PO4-(1.0M). Now we have to go HALF WAY through the second deprotonation.  This will be the half equivalence point. So to get to the 3rd acidic proton it will be another 10000 moles of H+, but we only want half, so that would be 5000.

Total acidic protons to get to the second half equivalence is 5,000 moles from HCl, 10,000 moles for first deprotonization of H3PO4, plus the half way of the 2nd deprotonization 5000 moles.

So total we need to neutralize 5,000+10,000+5,000 = 20,000 moles of acidic protons.

20,000 moles H+ ( 1 mole NaOH / 1 mole H+)( 39.997g NaOH / 1 mole NaOH)(1kg NaOH/1000g) = 799.94kg of NaOH.  Based on the sigfigs it only looks like 1 sigfig so it would be 800 kg

[Borek]

Logic behind your calculation looks more or less OK, but the number you are using are different from the numbers posted in the original question. 10,000 L or 1000L? 5 M or 0.5 M?

Technically you are right about the correct significant digits pointing to 300 kg, but to me 280 kg looks more reasonable. Sig figs are nasty sometimes.

[asmcriminal]

Logic behind your calculation looks more or less OK, but the number you are using are different from the numbers posted in the original question. 10,000 L or 1000L? 5 M or 0.5 M?

Technically you are right about the correct significant digits pointing to 300 kg, but to me 280 kg looks more reasonable. Sig figs are nasty sometimes.

Okay thanks fixed it. It's a misfortune I have that problem(The ADD) you can see how it affects me.   But yes, the sigfigs where unreasonable I believed, but to me that's what they where so I used them. But with the original numbers I had 280kg would have been better. 

[SweetAsC12H22O11]

That's a great explanation, thank you.

So I didn't even need V_e. Knowing how many moles of H⁺ need to be neutralized indicates how many moles NaOH are needed. Since HCl is a strong acid, it is neutralized first. Then, since the reaction goes to 1.5 V_e, H₃PO₄ H₂PO₄^- is complete, and half of H₂PO₄^-  HPO₄^2- is complete. Total moles H⁺ to be neutralized = 20,000.

Just needed to summarize for myself  Thanks!

[SweetAsC12H22O11]

I brought this problem up in study group today and found that a few people solved it differently, getting a different answer: 237 kg.

They calculated Ka1 and Ka2 from the pKa values and deduced that Ka1 > 100Ka2, so that the system can be treated at monoprotic.

Then, they calculated the concentration of H₃PO₄ using:

     Ka1 = x₂/F-x

But drop the x in the denominator since Ka1 > 100Ka2. X in the numerator represents the concentration of H⁺ and of OH^-. I'm assuming they used the concentration of OH^- to then calculate the mass of NaOH needed, adding the moles of HCl neutralized to the calculation.

I am now wondering why they attempted to solve it this way, does it seem reasonable? If not, why not?

[SweetAsC12H22O11]

Here is how I believe they solved the problem:

Ka1 = 7.1121 X 10^-3 & Ka2 = 6.3096 x 10^-8   

so Ka1>100Ka2   --> treat as monoprotic

Ka1 = ([H⁺][OH^-]) / F-x   

Ka1 = x²/F   --->   (7.1121 x 10^-3) = x²/1.0

Therefore x = 0.084333267 = [OH^-] x 10,000 L = 843.3326746 moles

     + 5,000 moles HCl = 5,843.332675 moles x 39.997 g/mol = 233,733.307

     / 1000 = 233.73 = 234 kg NaOH

Does the fact that Ka1>100Ka2 really mean we can approach the problem this way? I feel the original response posted here is more clear but just want to be sure...

[Borek]

No, you can't treat the acid as monoprotic. After neutralizing first proton stoichiometrically you end with a solution of an an amphiprotic salt, with pH well below 7. When you add more base, it will neutralize the second proton.

Besides, you (I mean plural here - your study group) should know what it means when pH=pKa. It says something important about the weak acid solution. Final pH of 7.2 happens to meet that condition.

[SweetAsC12H22O11]

Thank you!