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Topic: Equivalence point  (Read 4487 times)

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Offline Boxxxed

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Equivalence point
« on: February 25, 2011, 02:48:51 PM »
I graphed a titration curve for HCl and NaOH. Strong acid and strong base reached equivalence at 7 pH. (Seen from midpoint between tangents). I have the volume and concentration (moles) of NaOH added to the HCl solution.

I have been teaching myself as classes are weeks behind labs so some things are still fuzzy.

Do I just calculate the moles of NaOH at 7 pH from my data which will equal the moles HCl? (I have volume of HCl)

It is asking me for the concentration of the unknown HCl solution, so I think this makes sense

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Re: Equivalence point
« Reply #1 on: February 25, 2011, 03:14:19 PM »

At 7 pH approximately 24.35 ml of 0.1068 M/L NaOH was added.

The volume of the HCl to start with was 25 ml. Given it took an almost equal volume of NaOH to titrate the HCl solution the concentrations must be almost equal as well. I calculated the HCl concentration to be 0.11 M/L whic satisfies this.

I am still confused though, I have the initial pH of the HCl solution before adding any NaOH to be 1.14

Can't this be used to calculate the H3O+ ions and thus concentration of HCl?

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Re: Equivalence point
« Reply #2 on: February 25, 2011, 03:24:43 PM »
Do I just calculate the moles of NaOH at 7 pH from my data which will equal the moles HCl? (I have volume of HCl)

Yes, see

http://www.titrations.info/titration-calculation
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Re: Equivalence point
« Reply #3 on: February 25, 2011, 03:33:58 PM »
I am still confused with what exactly H3O ions are.

HCl+H2O--->H3O+Cl ?

so from initial pH of 1.14 of HCl solution couldn't I calculate

antilog-1.14= 0.072 moles/L? Why doesn't this equal approximately 0.1? Experimental error?

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Re: Equivalence point
« Reply #4 on: February 25, 2011, 03:56:36 PM »
so from initial pH of 1.14 of HCl solution couldn't I calculate

antilog-1.14= 0.072 moles/L? Why doesn't this equal approximately 0.1? Experimental error?

Yes. pH is -log of activity, not concentration, you should should take ionic strength of the solution into account. Electrode is usually calibrated for 4-7-10 points, around 1 you are outside of the calibrated zone. Plenty of reasons.
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Re: Equivalence point
« Reply #5 on: February 25, 2011, 04:41:38 PM »
CH3COOH titrated with NaOH (same concentration NaOH)

The volume of NaOH added was again close to 25 ml at equivalence, which is also the volume of the acid solution.

Is the concentration of the acid solution again similar to the concentration of the NaOH? Is there something I am missing with the conjugate base?

It is also asking for my % error of pKa. I derived the experimental value from my graphs but is there a numerical way to calculate this without the use of a graph? I'm not sure what they want


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Re: Equivalence point
« Reply #6 on: February 25, 2011, 05:17:58 PM »
CH3COOH titrated with NaOH (same concentration NaOH)

Just a stoichiometry, nothing else.

Quote
It is also asking for my % error of pKa. I derived the experimental value from my graphs but is there a numerical way to calculate this without the use of a graph? I'm not sure what they want

Part of the graph must be linear (see Henderson-Hasselbalch equation), you can fit a straight line to the points.
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