Acid-Base Titrations

CH 229 – General Chemistry 23. Acid-Base Titrations Introduction Acid base reactions are one of the most important chemical reactions there is. A common form of this type of reaction is called titration. Titration is the addition of a solution with a known concentration (the titrant) to a solution with a known volume but an unknown concentration (the analyte), you add the known titrant to the analyte till the reaction reaches neutralization. Titration is important in many ways but one specific way it impacts the real world is through our food. Titration allows food companies to determine the exact amount of salt in a certain food they are selling or to know the concentration of vitamin E (Bell-Young). Knowing these values ensures that the food is safe to eat. Titrations are important to chemistry because they allow you to discover the concentration of unknown solutions leading to new discoveries about how solutions react together. In this experiment we titrated a weak acid KHP with strong NaOH. This allowed us to find the molarity of KHP. We then applied this technique to a weak acetic acid solution and was able to calculqte the pH curve and finds its equivalence and half equivalence point. Balanced Chemical Equations (6 points, this includes that the equations are appropriate given the experimental results, which is also part of the discussion) ● Include any chemical equations as NET IONIC EQUATIONS with phase labels you used in the experiment. Do not use abbreviations for any of the substances, use chemical formulas. This should include at least ● The acid-base net ionic equation for the strong base/weak-acid titration occurring in the standardization 1. ࠵?࠵?࠵?(࠵?) + ࠵?࠵? − (࠵?࠵?) ⇌ ࠵? 2 ࠵?(࠵?) + ࠵?࠵? − (࠵?࠵?) ● The acid-base net ionic equation for the strong base/weak-acid titration that you carried out. 2. ࠵?࠵? 3 ࠵?࠵?࠵?࠵?(࠵?࠵?) + ࠵?࠵? − (࠵?࠵?) ⇌ ࠵?࠵? 3 ࠵?࠵?࠵? − (࠵?࠵?) + ࠵? 2 ࠵?(࠵?) ● The hydrolysis reaction (acid or base dissociation net ionic equation) that is responsible for the initial pH of the weak acid. 3. ࠵?࠵? 3 ࠵?࠵?࠵?࠵?(࠵?࠵?) + ࠵? 2 ࠵?(࠵?) ⇌ ࠵?࠵? 3 ࠵?࠵?࠵? − (࠵?࠵?) + ࠵? 3 ࠵? + (࠵?࠵?) ● The hydrolysis reaction (acid or base dissociation net ionic equation) that is responsible for the equivalence point pH in your acid/base titration. 4. ࠵?࠵? 3 ࠵?࠵?࠵? − (࠵?࠵?) + ࠵? 2 ࠵?(࠵?) ⇌ ࠵?࠵? 3 ࠵?࠵?࠵?࠵?(࠵?࠵?) + ࠵?࠵? − (࠵?࠵?) Results: Data, calculations, and graphs. Tables of Data and Calculation Results (10 points) Table 1: Titration of KHP with NaOH

Trial 1 2 3 Mass of KHP (g) .8805 .8811 .8680 Moles of KHP (mol) .004312 .004314 .004250 Volume NaOH 22.00 21.10 20.70 Molarity of KHP (mol/L) .1959 .2043 .2053 Avg Molarity of NaOH .2018 ± 0.00012 Avg Class Molarity NaOH .208 ± 0.016 Observations: The solution was clear and the KHP was a solid white looking salt that took some time for it to dissolve in solution. The titrant made a slight pink cloud in solution when added and faded until the solution passed the equivalence point which is when the solution went light pink. Table 2: Acetic Acid Titration Acid Acetic Acid Volume of Acid (mL) 10 Initial pH 3.01 Equivalence point Volume NaOH(mL) 12.7 Observations: The solution was a clear solution and when we added the titrant to the solution it turned it pink temporarily until the solution was at the equilibrium point where it turned a light pink and the more NaOH we added the darker the pink got. Table 3: Mandelic, Formic, Lactic and Acetic Acid Titrations Acid Mandelic Formic Lactic Acetic Molarity of Acid .229 ± .012 .220 ± .010 .230 ± .017 .261 ± 0.008 Initial pH 2.6 ± .7 2.7 ± .5 2.6 ± .3 2.8 ± .5 pKa 3.4 ± .6 3.6 ± .3 3.6 ± .3 4.5 ± .4 Ka 4. 0࠵?10 −4 2. 5࠵?10 −4 2. 5࠵?10 −4 3. 2࠵?10 −5 % dissociation 11 ± 10 8.1 ± 2.8 8.3 ± 3.3 3.7 ± 1.9 Calculations (8 pts) Part A

Figure 2: Partner Acetic Acid Titration Graph ● Overlaid titration curves for the four acids studied. Figure 2: Class Data Graph of Mandelic, Formic, Lactic and Acetic Acid Titrations Discussion What is the concentration of NaOH you determined ? (5 points)

The concentration of NaOH was determined by finding the moles of KHP in solution as seen in the part A calculations. We know that titration is putting in the same amount of NaOH or strong base in this case into a weak acid solution to neutralize it so once we determined the initial amount of moles of KHP we used equation #1 and changed to moles NaOH and divided that by L of NaOH used to get its molarity of 0.1959 M NaOH. The average was .2018 ± 0.00012 and the class average was .208 ± 0.016, which is very similar to what was calculated. What is the K a of the weak acid you measured? (5 points) The Ka of the weak acid was determined by finding the equivalence point where moles of NaOH was equal to acetic acid, then we recorded the volume of NaOH that it took to titrate acetic acid and looked at the titration curve that was graphed in the titration process and determined the pH at the equivalence point. Next you find the ½ equivalence point since according to the henderson hasselbalch equation at the half equivalence point pKa = pH. The ½ equivalence point is half the volume of the equivalence point and at that volume we found the pH to be roughly 4.5 which mean the pKa was 4.5 and the Ka is . These values are the same as what the class got as well and are very similar to the literature 3. 16࠵?10 −5 values with a pKa = 4.76 and Ka = . 1. 73࠵?10 −5 What is the relation between initial pH, percent dissociation, and acid strength (K a )? (5 points) ● Answer this using class data and make sure to follow the model of making specific reference to the class data, discussing what you expect based on your understanding of chemical phenomena (i.e. your understanding of general chemistry), and discussing any disagreements between the class data and the expectations based on chemical phenomena. The relationship between initial pH, percent dissociation and acid strength can be seen by looking at the class data of acetic and mandelic acid. Mandelic acid has a lower pH of 2.6 ± .7 which shows how it dissociates better in solution as seen in its percent dissociation of 11% ± 10. Mandelic acid also has a higher Ka value of . This makes sense since K = products / reactants and if Mandelic acid is 4. 0࠵?10 −4 dissociating better than acetic acid that would mean it’s forming more products (hydronium) therefore having a higher Ka. Acetic acid has a higher pH of 2.8 ± .5, and a lower % dissociation of 3.7% ± 1.9 followed by a smaller Ka value of . It can be seen through the class data that as pH lowers the 3. 2࠵?10 −5 % dissociation increases and the Ka increases as well because of the increase in Hydronium and conjugate base concentration in solution and decrease of weak acid as seen in equation 3. How does the shape of the titration curve change as the acid strength (or Ka) decreases? (5 points) The titration curve becomes less steep and the equivalence pH increases as the acid strength decreases. In figure 2 it can be seen how acetic acid has a shorter jump in pH at the equivalence point than lactic acid does, and the overall titration curve is shifted slightly higher than the other acids. In table 3 it can be seen how acetic acid has a smaller Ka value than lactic acid, which shows that acetic acid is a relatively weaker acid than the other acids. When you look at the graph these trends of the lines make sense. They make sense because since acetic acid is weaker than the other acids its conjugate base would be stronger so once NaOH is introduced into the solution this stronger conjugate base thats being formed causes the line