Title Experiment 2: Acid-Base Reactions: the Identification of an Unknown Weak Acid

Transcription

1 Title Experiment : Acid-Base Reactions: the Identification of an Unknown Weak Acid Name Manraj Gill (Lab section 1) Introduction The end goal of the set of experiments and analysis performed in this lab is to determine the identity of an unknown acid. Since we do not directly obtain a solution of sodium hydroxide of a known concentration, the first step (Part I of Experiment ) is focused on standardizing a solution of sodium hydroxide that we made and trying to accurately determine its concentration. In Part II, we use this standardized solution to titrate an unknown weak acid. This titration, measured using a meter (potentiometric titration), allows us to identify chemical properties of the acid such as the molecular weight and pka because the measurements can accurately discriminate between a monoprotic and a diprotic acid. Data and Results Part I Indicator Titration The indicator titration allows us to determine the concentration of the NaOH to some rough approximation that will then be helpful in taking precise measurements in the potentiometric titrations. The titration relies on phenolphthalein indicator that changes from colorless to pink at the endpoint. Based on the in-lab work shown on page 9 of the attached notebook pages,.mmol of the KHP (potassium hydrogen phthalate) reached the end point when 5.ml of the NaOH solution was added. The end point (when the solution stabilizes at pink for >3 seconds) is reached when equal number of moles of H + and OH - are present in solution. Therefore, at the equivalence point,.mmol of OH - is in solution. This gives us a rough estimate of the concentration of NaOH:.mmol in 5.ml =.79M Using this information, we can predict before titrating, the volume of the NaOH that will be required if the mass of the KHP is known in the potentiometric titrations.

2 Potentiometric Titrations Potentiometric Titrations Mass of KHP.3315g.5g.951g Millimoles of KHP Predicted End Point (based on Indicator Titration) (in milliliters) NaOH NaOH NaOH

3 Analysis of Potentiometric Titrations can be done with graphical analysis as the raw data of measurements can be translated into a titration curve that almost explicitly highlights the equivalence point. An estimate of the equivalence point in each of the 3 potentiometric titrations will help us calculate the molarity of the NaOH solution to a greater certainty. Potentiometric Titration of NaOH Added (milliliters) Potentiometric Titration of Linear Regression (Before Eq) Linear Regression (After Eq) Joining the Midpoints Maximum Slope Linear (Linear Regression (Before Eq)) Linear (Linear Regression (After Eq)) Linear (Joining the Midpoints) Linear (Maximum Slope) Equation of Before Eq y =.97x +.3 Equation of After Eq y =.13x +.53 Average of y-intercept.55 Average of slope.1135 Equation of Line Joining Midpoints of Before and After Eq y =.1135x Equation of Max. Slope y =.795x x-coordinate at the point of intersection of the two lines 1.ml NaOH This gives us.75m as the concentration of NaOH. But to confirm this value, we perform the same calculation of the two other sets of data to get an agreement within.5% of the mean value.

4 Potentiometric Titration of Potentiometric Titration of 3 Linear Regression (Before Eq) Linear Regression (After Eq) Joining the Midpoints Maximum Slope Linear (Linear Regression (Before Eq)) Linear (Linear Regression (After Eq)) Linear (Joining the Midpoints) NaOH Added (milliliters) Linear (Maximum Slope) x-coordinate of the intersection: 31.9ml NaOH Therefore,.7733M NaOH 1 1 Potentiometric Titration of 1 Potentiometric Titration of Linear Regression (Before Eq) Linear Regression (After Eq) Joining the Midpoints Maximum Slope Linear (Linear Regression (Before Eq)) Linear (Linear Regression (After Eq)) Linear (Joining the Midpoints) NaOH Added (milliliters) Linear (Maximum Slope) x-coordinate of the intersection: 31.3ml NaOH Which gives,.7737m NaOH Mean value of molarity:.79m with a standard deviation of. Giving us a 95% confidence interval of.1!

5 Part II Since my partner from Part I was not present during Part II, collaborated with Tanner Adams and Nick. Tanner Adams NaOH solution was used and it had a concentration of.91m As shown in calculations performed in lab (shown in the attached lab notebook pages), the amount of unknown acid to be used was calibrated to yield an endpoint at ~5mls. This was determined based on an initial indicator titration performed using a higher amount of the unknown. Unknown sample number: #17 It was determined that.7g of the unknown would be required and.9g of NaCl to create a solution with.1m ionic strength. Raw data from the potentiometric measurements on the next page (in order to fit it all in a single page) Analysis of the data: 1 Potentiometric Titration of 1 1 Potentiometric Titration of 1 Linear Regression (After Eq) Linear Regression (Before Eq) Joining Midpoints Maximum Slope Linear (Linear Regression (After Eq)) Linear (Linear Regression (Before Eq)) NaOH Added (ml) Linear (Joining Midpoints) Linear (Maximum Slope) Based on just the first titration, it was evident that there was only one equivalence point observed and the second one could not be determined since it seems to lie too far up the scale. The x-intercept of the intersection occurs at.15ml of ~.9M NaOH. Performing similar calculations on the other two data (Titration of and 3) will give us a better approximation of the true first equivalence point which can be used to determine the Ka(1)!

6 Potentiometric Titrations 1 3 NaOH NaOH NaOH

7 Continuation of the analysis: 1 Potentiometric Titration of 1 Potentiometric Titration of After Eq Before Eq Midpoint Max Slope Linear (After Eq) NaOH Added (ml) Linear (Before Eq) Linear (Midpoint) Linear (Max Slope) x-intercept:.1ml NaOH 1 Potentiometric Titration of y = 9.95x Potentiometric Titration of 3 After Before Midpoint Max Slope Linear (After) NaOH Added (ml) Linear (Before) Linear (Midpoint) Linear (Max Slope) x-intercept: 1.ml NaOH Average value of the NaOH amount at the first equivalence point:. with a standard deviation of.15ml Therefore, since.91m NaOH was used, the moles at the first Ka corresponds to.33millimoles

8 The estimate of the first Ka is expected to be at half of the equivalence, therefore at a of. and since at the half equivalence point, pka =, pka(1) of weak acid #17 is ~.. Based on this information and the knowledge that the second equivalence point is not reached, we can conclude that the identity is Iminodiacetic Acid!