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CHEM-002 Lecture Notes - Lecture 19: Calcium Fluoride, Equivalence Point, Titration

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chocolatedeer31
20 Mar 2017
School
Georgetown University
Department
Chemistry
Course
CHEM-002
Professor
Dedios Angel

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Document Summary

Any titration problem is going to include the half-equivalence point, at which point the ph = pka. This is when you have as many moles of strong base as you do of the weak acid. In a polyprotic acid, you do it like there are two acids, because the eqvilance points are so different. This goes to completion, then ha- reacts with oh- to form a2-. So in these, he half equilance point is very close to pka1. You also do(cid:374)"t have the thing where oh = sqrt(kb1 x c) because ha- is both an acid and a base, so you have to consider both. Si(cid:374)(cid:272)e they"(cid:396)e fo(cid:396)(cid:373)ed at the sa(cid:373)e ti(cid:373)e, thei(cid:396) (cid:272)o(cid:374)(cid:272)e(cid:374)t(cid:396)atio(cid:374)s ha(cid:448)e to (cid:271)e e(cid:395)ual at the fi(cid:396)st equivalence point. H2a a2- + 2h+ has a k (cid:448)alue of ka(cid:1005)ka(cid:1006) (cid:271)e(cid:272)ause it"s a (cid:272)o(cid:373)(cid:271)i(cid:374)atio(cid:374) of those t(cid:449)o (cid:396)ea(cid:272)tio(cid:374)s.

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Related Questions

Could someone please help me with this chemistry problem? Thank you!

After performing a titration using 20 mL of unknown weak acid with 0.3830M NaOH, I obtained the following results:

0 mL added NaOH = pH 3.80

Final amount of added NaOH = 8.59 mL = pH 12.20

At Equivalence Point: Volume of NaOH = 4.835 mL; pH = approx 8.40

At the Halfway Point; Volume of NaOH = 2.418 mL: pH = 5.15

Ka = 7.08 x 10^-6

Initial [HA] = 0.0926M (determined by M1V1=M2V2 at the equivalence point)

Molar Mass = 1079.8 g/mol (determined by using original mass 1.9977 g/1.85 x 10^-3 moles

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Now I am asked the following questions in order to check my results for consistency;

Using your experimental Ka value, the concentration of your unknown acid solution and the concentration of the standardized NaOH solution, CALCULATE the pH of the following points on your titration curve:

1) The beginning of the titration before any any NaOH was added

2) 25% of the way to the equivalence point (25% of 4.835 mL is 1.21 mL)

3) 75% of the way to the equivalence point

4) The equivalence point (remember, this is found using Ka and concentrations of acid and base)

5) 20% of the way to the equivalence point (1.20 x 4.835= 5.80 mL)

I've found some info on the internet that says when in the buffer region, the moles of A- = moles of base added and that moles of HA = original moles of HA - moles of base added. Where I got confused was when the equilibrium expression was written as Ka = [H+] [mols Base Added/Volume] ÷ [mols Acid Left/Volume]. I don't know what volumes to use because I always think it makes more sense to use total volume of acid plus base mL's. But maybe you are supposed to use only the volume of each (acid/base) separately, in which case I think the volume of acid would be 20 mL. Maybe this is a totally wrong way to do it- I don't know! Please Help!!!!

1. The accepted dissociation constant for acetic acid is 1.8 x 10-5. One student's graphical determination of the Ka of this acid is 1.4 x 10-5. Calculate the percent error in this student's determination.

2.The equations used in this experiment would have to be modified slightly if an acid with a Ka of 1.0 x 10-3 or greater were used as an unknown. Assuming such a situation, explain the problems that would result from the use of the equations in their present form. What modifications would you make to give accurate results when using such an unknown?

a. The following two reactions happen in the solution. OH- + HAn LaTeX: \longrightarrow An- + H2O (Reaction 1) HAn LaTeX: \longrightarrow An- + H+ (Reaction 2) Within this lab, the acid used is a very weak acid with a Ka much smaller than 1 x 10-3. So the dissociation of the acid (Reaction 2) is negligible. The moles of An- at equilibrium is calculated as equal to the initial moles of OH-, and mole of HAn at equilibrium is calculated as the initial moles of HAn minus the initial moles of OH-.

b. When the Ka of the weak acid is equal or bigger than 1 x 10-3, the dissociation of acid (Reaction 2) is no longer negligible. In such case, An- is produced from both Reaction 1 and Reaction 2. The moles of An- produced from Reaction 1 is equal to the initial moles of OH-; and the moles of An- produced from Reaction 2 is equal to the moles of H+ (or H3O+), which can be calculated as molarity of H+ times liters of the solution mixture. So the moles of An- at equilibrium should be calculated as: moles of An- at equilibrium = initial moles of OH- + moles of H+ ; (Equation 1) And the moles of HA at equilibrium should be calculated as: moles of HA at equilibrium= initial moles of HA -moles of An- at equilibrium (Equation 2)

c. If we keep ignoring Reaction 2 even though the Ka of the weak acid is equal or bigger than 1 x 10-3, we will end up with an incorrectly smaller mole and smaller equilibrium concentration of An- and incorrectly bigger mole and bigger equilibrium concentration of HAn , thus an incorrectly smaller Ka. The problem can be avoided and an accurate determination of Ka can be obtained by calculating moles of An- at equilibrium and that of HAn by Equation 1 and Equation 2, respectively.

d. The following two reactions happen in the solution. OH- + HAn LaTeX: \longrightarrow An- + H2O (Equation 1) HAn LaTeX: \longrightarrow An- + H+ (Equation 2) Within this lab, the acid used is a very weak acid with a Ka much smaller than 1 x 10-3. So the dissociation of the acid (Equation 2) is negligible. The moles of An- at equilibrium is calculated as equal to the initial moles of OH-, and mole of HAn at equilibrium is calculated as the initial moles of HAn minus the initial moles of OH-. When the Ka of the weak acid is equal or bigger than 1 x 10-3, the dissociation of acid (Equation 2) is no longer negligible. In such case, An- is produced from both Equation 1 and Equation 2. The moles of An- produced from Equation 1 is equal to the initial moles of OH-; and the moles of An- produced from Equation 2 is equal to the moles of H+ (or H3O+), which can be calculated as molarity of H+ times liters of the solution mixture. So the moles of An- at equilibrium should be calculated as: moles of An- at equilibrium = initial moles of OH- + moles of H+ ; (Equation 3) And the moles of HA at equilibrium should be calculated as: moles of HA at equilibrium= initial moles of HA -moles of An- at equilibrium (Equation 4) If we keep ignoring Equation 2, we will end up with an incorrectly smaller equilibrium concentration of An- and incorrectly big equilibrium concentration of HAn , thus an incorrectly smaller Ka. The problem can be avoided and an accurate determination of Ka can be obtained by calculating moles of An- at equilibrium and that of HAn by Equation 3 and Equation 4 respectively.

e. all are correct

Sketch the titration curve for a weak acid titrated with a strong base.

 

When performing calculations concerning weak acid-strong base titrations, the general two-step procedure is to solve a stoichiometry problem first, then to solve an equilibrium problem to determine the pH. What reaction takes place in the stoichiometry part of the problem? What is assumed about this reaction?

At the various points in your titration curve, list the major species present after the strong base (NaOH), for example reacts to completion with the weak acid, HA. What equilibirum problem would you solve at the various points in your titration curve to calculate pH? Why is pH at> 7.0 at the equivalence point of a weak acid-strong base titration? Does the pH at halfway poin the equivalence point have to be less than 7.0? what does the pH at the half-way point equal? Compare and contrast the titration curves for a strong acid-strong base titration and a weak acid-strong base titration.