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Brian Wahlberg Chemistry 113 Mike Kelson – H5 November 23, 2015

Determination of Ka: Titration of a Weak Acid

INTRODUCTION Every living creature’s body survives in a constantly changing internal environment, as the organism works around the clock to maintain a balance in their system. Both acidic environments and basic environments coexist within organisms’ bodies. Acids have high concentrations of H+ ions, meaning they have a very low pH, while bases have a very low concentration of H+ ions and high levels of OH- ions, resulting in a very high pH. When acids and bases combine, depending on the strength of each, the two will slowly escalate or decline to a higher or lower pH respectively. This is based off the strength of both the acid and the base. A weak acid will not win the battle with a strong base, so the pH at the equivalence point will be above the neutral point of 7. The equivalence point is the volume of either the acid or base added in which the moles of both are equal in the solution. Here on the graph we will see a large spike in pH with a very minute amount of solution added. An extremely important aspect of any acid in these reactions is the ionization constant, Ka, which we will be experimentally solving for in the titration of CH3COOH with NaOH. PROCEDURE First we set up the apparatus which included a 60 mL reagent reservoir, a drop counter, a stir station with a stir bar and the pH probe. Next we set up our computer by hooking up the drop counter and pH probe to the interface and by opening up the Logger Pro program with the

correct file. Once we were set up, we tested the reservoir for leaks by filling it with water, and then filled it with the 0.100 M NaOH solution. By messing with one valve while leaving the other completely open, we obtained a drop rate of about one drop per second. Then we calibrated the drop volume for the drop counter using Logger Pro so it knew how many drops fell per milliliter. After, we placed a 100 mL beaker of 20 mL of acetic acid on the stir plate and put the pH sensor in the slot in the drop counter so that it hangs within the acid in the beaker. We turned on the magnetic stirrer to a moderate rate and started collecting our data. We opened the valve that didn’t control the rate and watched our data start to plot. Then we waited until our graph had reached the equivalence point and then leveled out a bit to stop collecting data. With our plotted data, we opened Page 3 showing the second derivative and found the highest peak which gave us the point of inflection and, therefore, our equivalence point. Finally, we disposed of our beaker contents and cleaned up our station. EQUATIONS

Ka =

+¿¿ H ¿ −¿ ¿ A ¿ ¿ ¿

pH (1/2) = pKa

DATA COLLECTION

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pH of original acid without base added = 3.00 Molarity of [NaOH] = .0982 Moles/Liter Equivalence point = 40.66 mL CALCULATIONS Original Concentration of CH3COOH 40.66 mL ∗1 L 1 ∗.0982 moles NaOH 1000 mL ∗1 mole C H 3 COOH 1L ∗1 1 mole NaOH ∗1000 mL 20.0 mL 1L

= 0.1996 M

Determination of Ka through two methods Using pH of original acid solution; pH = 3.00 so therefore -log10[H+] = 3.00

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[H+] = 0.001 M = [A-] +¿¿ H ¿ [.001 ]∗[ .001] −¿ Ka = Ka = = 5.01*10-6 ¿ [.1996 ] A ¿ ¿ ¿ Using the pH at half equivalence point; pH where 20.33 mL added = 5.16 pH (1/2) = pKa (5.16) = -log10Ka 10-5.16 = 6 . 9 1 8* 1 0-6 DISCUSSION/CONCLUSION Once the titration was complete and we had a nice titration curve, we found our equivalence point by looking at the highest point on the first derivative. We had two volume points, so we averaged them to get the volume of NaOH added at our equivalence point which was 4 0.66mL. The pH at this point was 9.52, which is corroborated by the original titration curve. With this volume, we found the moles of NaOH in our .0982M solution. The moles of NaOH and moles of acetic acid are the same due to their one to one ratio, so we divided this by 20 mL which was our original amount to get the molarity. Our original measurement of the pH of acetic acid was 3.00, so using this we found the concentration of hydrogen ions and also the conjugate base. We then plugged in our values into the Ka equation for the dissociation of a weak acid and got 5.01*10-6 as our value. For the second method of finding our Ka value, we looked at the data section and found what the pH was at half the volume of our equivalence point. At 20.33 mL, the pH was 5.16. This is also equal to the pKa value. Raising ten to the negative value of our pH, we found the Ka to be 6.918*10-6. We cannot calculate percent error without the real value of the ionization constant, however, there were definitely sources of error in our experiment that altered our final values. These sources could be not measuring out the acid precisely, the drop counter calibration, and rounding in our calculations. All these sources could possible have affected our calculated Ka values. This experiment was successful in demonstrating the two methods to calculate the acid dissociation constant. Our titration curve was great and we got a basic final solution, which was expected. If I were to do this experiment again, I would just try to minimize the sources of error.

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