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Topic: Hardness-Titration  (Read 2100 times)

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Offline serotonin

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Hardness-Titration
« on: May 03, 2024, 02:24:10 PM »
Please help?!:/
We titrate a solution containing calcium ions (Ca2+), V=100 mL, with EDTA solution 0.1 M. The end point is determined at 10 mL. What is the equivalent point of the process? Based on the reaction, calculate the total hardness of the sample in mol/L CaCO 3 and mg/L CaCO3. Calculate the content of calcium ions (Ca2+) as [Ca2+] and the contents [EDTA ] and [Ca2+-EDTA ] after adding 5mL of titrant. Explain based on reactions.

So far I've tried: nEDTA=(0.1*0.01)=0.001mol, nCaCO3=0.001mol (reaction 1:1)
Conversion: mCaCO3=n*molar mass=0.10009g then mg/L=mCaCO3*VH2O*1000mg/g=1000,9mg/L

Then, after adding 5mL of titrant: cCaCO3=molesCa2+/VH2O=0.0005*0.105L=0.00475mol/L,
For the equivalent point, I thought of using cAvA=cBvB => (0.1)(0.1)=cB(0.01) but Im not sure.
For EDTA concentration" (10-5mL)=> nEDTA/vH2O=(0.1mol/L * 0.01L * 0.0005mol)/(0.01L-0.005L)=0.1mol/L
Ca-EDTA: 0.1mol/L - 0.00476mol/L = 0.09524mol/L
(Not sure) should volume after adding be 100mL+5mL?

Any help is MUCH appreciated. Thank you.

Offline Borek

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Re: Hardness-Titration
« Reply #1 on: May 03, 2024, 03:12:13 PM »
So far I've tried: nEDTA=(0.1*0.01)=0.001mol, nCaCO3=0.001mol (reaction 1:1)
Conversion: mCaCO3=n*molar mass=0.10009g then mg/L=mCaCO3*VH2O*1000mg/g=1000,9mg/L

OK

Quote
Then, after adding 5mL of titrant: cCaCO3=molesCa2+/VH2O=0.0005*0.105L=0.00475mol/L,

That's a reasonably good answer.

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For the equivalent point, I thought of using cAvA=cBvB => (0.1)(0.1)=cB(0.01) but Im not sure.

No. You need stability constant here.

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For EDTA concentration" (10-5mL)=> nEDTA/vH2O=(0.1mol/L * 0.01L * 0.0005mol)/(0.01L-0.005L)=0.1mol/L

No. EDTA reacted with the Ca2+, so its concentration is much lower.

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Ca-EDTA: 0.1mol/L - 0.00476mol/L = 0.09524mol/L

Close, but you forgot the dilution (total concentration of Ca in both complexed and free form is no longer 0.1M)

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(Not sure) should volume after adding be 100mL+5mL?

Yes, and most your calculations already take this into account, just not all.

compare http://www.titrations.info/titration-curve-calculation
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Offline serotonin

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Re: Hardness-Titration
« Reply #2 on: May 03, 2024, 04:07:16 PM »
Thank you so much for replying, I appreciate it.

I'm not sure about EDTA concentration. Should I say: nEDTA/volume=(0.1mol/L*0.01L*0.0005mol)/0.105L instead?

For equivalent point, it gives Kf=1*10^10. I thought saying: Kf= (CaEDTA2-)/((Ca2+)*(EDTA4-))=1*10^10 then put 0.1M for both Ca2+ and EDTA. But I don't think that's correct.

Would there be any chance of guiding me a bit about Ca-EDTA concentration? If not, completely understandable.
Thank you again SO much for your help.

Offline Borek

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Re: Hardness-Titration
« Reply #3 on: May 03, 2024, 05:36:04 PM »
For half titration you should assume EDTA reacted with Ca2+ in the solution completely, and plug known concentrations into Kf, the only unknown will be the EDTA concentration.

That's only an approximation (to get the more exact value you should use ICE table), but typically gives acceptable accuracy.

It won't work for equivalence point though. There you can assume you have solution equivalent to the one prepared by dissolving just the proper amount of CaEDTA and calculating how much of it dissociated (hint: concentrations of Ca2+ and unreacted EDTA will be identical). Again, use Kf.
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Offline serotonin

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Re: Hardness-Titration
« Reply #4 on: May 03, 2024, 07:57:08 PM »
I'm sorry for asking this again, I'm not good with titrations and my brain is stuck. Should I do half titration as well? For equivalence would: [CaEDTA] = 10^10 * (0.01 mol/L)^2 be somewhat correct? as Ca2+ and EDTA are identical.
(Given that the volume of the solution titrated is 100 mL (V=100 mL) and the amount of EDTA solution added is 10 mL, the initial concentration of EDTA can be calculated as:
Initial concentration of EDTA = (0.1 mol/L) * (10 mL / 100 mL) = 0.01 mol/L) I know it's same as before, but for some reason I can not get it. Thank you very much for your help. SO SO appreciative

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Re: Hardness-Titration
« Reply #5 on: May 04, 2024, 01:50:45 PM »
I'm sorry for asking this again, I'm not good with titrations and my brain is stuck. Should I do half titration as well?

Question, as stated in you first post, says

Calculate the content of calcium ions (Ca2+) as [Ca2+] and the contents [EDTA ] and [Ca2+-EDTA ] after adding 5mL of titrant.

5 mL is half of 10mL required to get the endpoint, so you are asked to calculate things at halfopint.

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For equivalence would: [CaEDTA] = 10^10 * (0.01 mol/L)^2 be somewhat correct?

No. As I wrote, at the equivalence you have solution that is identical with one prepared by dissolving 0.01 moles of CaEDTA in 110 mL of water. 0.01 moles, as you mixed Ca and EDTA in stoichiometric amounts - there were 0.01 moles of EDTA added, so there were 0.01 moles of Ca present.

Some of CaEDTA will dissociate, you can calculate concentrations after dissociation using ICE table.
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Offline serotonin

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Re: Hardness-Titration
« Reply #6 on: May 05, 2024, 01:52:50 PM »
Thank you so much I really appreciate it.

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