Lab 2 - Determination of Ka of an Unknown Weak Acid: Acid-Base Equlibria During Titrations PDF

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Determination of Ka of an Unknown Weak Acid: Acid-Base Equlibria During Titrations...

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Determination of Ka of an Unknown Weak Acid: Acid-Base Equlibria During Titrations Purpose The intention of this experiment was to a) learn how to use volumetric glassware for the titration of an unknown weak acid with 0.1010M NaOH and b) to learn how to use a titration curve and pH meter to calculate the equilibrium constant (Ka) of a weak acid. This experiment was also used to learn about natural behavior of titration curves and the inherent data they provide.

Theory The theory behind this experiment lies in the behavior of titration curves and the data they provide in learning about acid-base equilibria and acid-base chemical behavior. In this specific acid-base reaction, an unknown concentration of weak acid reacts with a known concentration of strong base the titration curve will sharply rise with the initial amounts of strong base added, and then the pH will plateau as more strong base is added. However, when the amount of strong base begins to reach the amount of strong base needed to completely react with the weak acid, the curve begins to increase at a nearly vertical slope as this is the region when the equivalence point is reached. The equivalence point is the point where the amount of moles of base added is equivalent to the amount of moles of acid in the unknown amount of acid. The curve then begins to level off once again at a pH above the pH of the equivalence point as the addition of base makes the solution more basic. In this reaction between strong base and weak acid, the equivalence point is said to lie beyond pH 7 as the weak acid acts as a buffer against the addition of the base: That is until the buffer capacity of the weak acid is reached, which is the point where the equivalence point is reached. The equilibrium constant of the weak acid can be calculated in two ways: through the use of the equilibrium constant equation or through the use of the Henderson-

Hasselbalch equation. The first method is as follows

+¿¿ H ¿ −¿ ¿ OH ¿ ¿ K a=¿

in which before the base is

added the [H+] equals the [OH-] and the initial pH can be used to calculate the [H+] and this value can be inputted into the equilibrium constant equation to derive the Ka. The −¿ ¿ A ¿ latter method, the Henderson-Hasselbalch equation ( ), can be ¿ pH= p K a +log ¿ manipulated to find the Ka by using the half-equivalence point found from the titration. When the half equivalence point is used, [A-] equals [HA] which gives the log of these two a value of 0 so the new equation looks like

p H 1 /2 =p K a . This method can be

used by finding the volume of the equivalence point taking half of that volume and finding the pH of the halved volume, the volume of the half-equivalence point.

Procedure This titration was conducted twice for the solutions of #309 and #311. However, the general titration was the same. Before titrating, prepare a dilute solution of the unknown acid by taking 10mL of distilled water and mixing that with 10mL of unknown acid and putting that in a 100mL beaker with a magnetic stirring rod. The 10mL are measured using a volumetric pipet to ensure accuracy. Then the pH is tested of this prepared solution and recorded for future use. Then the buret is filled a sufficient amount of 0.1010M NaOH and the volume of this is recorded. Then the 100mL beaker of unknown solution is placed on a magnetic stirring platform, which is under the buret. The tip of the buret is lowered enough to prevent loss of base from splash, and the magnetic stirring platform is turned on and set to a low amount to ensure even stirring of mixing solutions. Then the pH meter is submerged in the solution so as it is not touching the bottom or is too close to the stirring rod. Once this set up is complete, the buret is opened to allow increments of about 0.2 - 0.5mL.While this is occurring, the investigator must

record the volumes and the resulting pH values. Once the titration nears a pH of 5, the investigator should stop titrating allow the solution to reach equilibrium and after that has been established, then begin to add 1 to 2 drops of base as the equivalence point is reached, and the investigator should notice the sharp increase in pH. While this is occurring, the pH values should be recorded before the volume as the pH will allow the investigator to see if the equivalence point has been reached. Once the equivalence point has been reached, the pH will begin to level off and then the investigator can add about 2 to 5 mL until the pH of 12 is reached. Upon completion of both titrations, the pH meter should be cleaned and put back in its calibration solution, the beakers, the burets, and the pipets should be cleaned and the equipment returned to their respective places.

Raw Data

Processed Data

Using the second derivative graph, the titration curve can be used to uncover the equivalence point which we will use in the calculation for Ka. Equivalence Volume for Solution #309 is the average of the two highlighted points (at these points the second derivative crosses the x-axis meaning there is an inflection point in the original graph and this inflection point is the equivalence point) so the equivalence volume is 11.825mL. And the pH at this point is 7.90 (calculated by the average of the two highlighted volumes’ pH values). Equivalence Volume for Solution #311 is the average of the two highlighted points (at these points the second derivative crosses the x-axis meaning there is an inflection point in the original graph and this inflection point is the equivalence point) so the equivalence volume is 11.530mL. And the pH at this point is 8.13 (calculated by the average of the two highlighted volumes’ pH values).

Using Ka equation

Solution #309:

+¿ ¿ H ¿ ¿2 & [HA]= ¿ ¿ K a=¿

1 0.1010mmol NaOH × 11.825m L × =0.0597 M mL 20.00 mL

So

K a=

(10−2.49 )2 −4 =1.7540 ×10 0.0597

Solution #309:

+¿ ¿ H ¿ ¿2 & [HA]= ¿ ¿ K a=¿

1 0.1010mmol NaOH × 11.530mL × =0.0582 M mL 20.00 mL −2.47 2

So

K a=

( 10 ) −4 =1.9728 ×10 0.0582

Using Henderson Hasselbalch Equation Solution #309:

p H 1 /2 =p K a and

p H 1 /2 =3.61 therefore

K a=10−3.61=2.4547 ×10−4 Solution #311: K a=10

−3.65

p H 1 /2 =p K a and

=2.2387 ×10

p H 1 /2 =3.65 therefore

−4

Discussion The processed data shows us that the theory that weak acid and strong base titrations result in an equivalence pH value being higher than 7 as the graphs are shifted to the right and up. The processed data also shows the buffer behavior of a weak acid. The two methods for solving for the equilibrium constant reveals two distinctly different values for equilibrium constant for both solutions. For example in solution #309, the difference between the values are

7.007 ×10

−5

and while this may not seem great the

fact is there seems to be a significant difference. However, the behavior of a weak acid is still revealed in both the equilibrium constant values as they show the fact that the weak acid does not readily dissociate thereby a small value of Ka. The error from this has probably arisen from the fact that the equivalence volume was not calculated in the best manner possible as a program that could interpolate the equivalence pH would result in a

more accurate value in the Henderson-Hasslebalch equation solving. The error probably also resulted in an inaccurate value for the equivalence volume which would in turn affect the value for the Ka when calculated in the equilibrium constant method. Another error that affected the results was the method of dropping base into the dilute acid solution. By correcting this error, more data could be collected of the change in pH versus mL of base added which would in turn provide a more accurate graph to interpolate the equivalence value. A way to completely eliminate this error would be to use a Vernier drop counter that has been previously calibrated in order to get an accurate account of the amount of base added, and then the LoggerPro software that is used in addedndum to this drop counter would provide the most accurate value for the equivalence point, and thereby the two methods of finding the equilibrium constant for the acid would be the most accurate. However, based on the theory, this experiment proved the following points: a) weak acids do not dissociate as readily as strong acids, b) they have small acid equilibrium constants and c) the equivalence point lies above pH 7 for a titration between a strong base and a weak acid....