CHEM237L Lecture Notes - Titration Curve, Equivalence Point, Titration

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purplemosquito736

15 Jul 2014

School

University of Waterloo

Department

Chemistry

Course

CHEM237L

Professor

Laura Marrone

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Document Summary

Titration curve of unknown amino acid #34 - trial 1. Buffer point at which excess hcl is neutralized (0. 946 mmol) Titration curve of unknown amino acid #34 - trial 2. Zone point at which excess hcl is neutralized (1. 088 mmol) First derivative plot for trial 1 d e d d a. 4. 726 mmol oh- d e d d a o m m. 8. 0 8. 2 8. 4 8. 6 8. 8 mmol oh added. 2nd equivalence point is at 8. 506 mmol oh- Sample calculations: to get mmol of oh- added, to get ph/ mmol oh added, determining pka values: To determine the ph, the value was estimated by looking at the ph when that amount of mmols is added. An average of the two ph values is taken to result in the experimental ph value. To calculate the first pka value, we know that both pka"s are equal distance from the first. Q-1. (a) the pka values were determined directly from the titration curves.

Related Questions

For your next titration you weigh out an actual amount of 624.3 mg of your sample. Dissolving in water and then titrating with 0.1482 M NaOH produces the pH titration curve shown below:

1.)Based on the above curve, what volume of NaOH (mL) was used to reach the equivalence point ?

Enter a numeric answer only, to three significant figures

2.)How many mmol of NaOH were added to the 624.3 mg of weak acid sample in order to reach the equivalence point?

3.)How many mmol of weak acid are in the titrated sample ?

4.)The molar mass of a compound is calculated by adding up the molar mass of the individual elements. This can be done when you know the molecular formula of a compound.

The effective molar mass (EMM) is used in identifying an unknown compound. A sample of the unknown is weighed to find the mass. Then an analytical technique is used to find the number of moles contained in the sample. Here the method used is titration of the weighed sample with NaOH of a known molarity. Dividing the mass of the sample by the moles found from the titration gives the EMM.

What is the EMM of the 624.3 mg sample of unknown acid that was titrated?

5.)What volume of NaOH was added to reach the half-titration (or half-equivalence) ?

6.To two significant figures, what is the pK_a of the unknown acid?

Please explain. Regarding titrations:

PART A:

Before the first endpoint only K₃PO₄ is reacting.

PO₄^3â_(aq) + H₃O⁺_(aq) D HPO₄^2â_(aq) + H₂O _(l)

If at the first equivalence point 12.93 ml of 0.1342 M HCl is consumed, how many mmol of K₃PO₄ were present in the sample? Round your answer to 3 decimal places. Units are not required.

PART B:

For the diprotic titration you will be performing in the lab your unknown solution will be a mixture of PO₄^3â and HPO₄^2â. At the first equivalence point all of the PO₄^3- will have been converted to HPO₄^2-, so the solution will now contain only HPO₄^2â . In a similar calculation to that in the previous question, a student has found that their original solution contained 1.352 mmol of PO₄^3-. If reaching the second equivalence point consumed 20.11 mL more of 0.1185 M HCl, how many mmol of HPO₄^2â were in the original sample? Report answer to 3 decimal places. No units necessary.

You have 1 litre of a 10 mM acetic acid/acetate buffer at pH 4.8. Will this buffer be able resist a change in pH when 8 mmol of HCl or NaOH is added? The approximate pKa value for acetic acid is 4.8.

CHEM237L Lecture Notes - Titration Curve, Equivalence Point, Titration

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