Titration diprotic acid PDF

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Titration of a Diprotic Acid Lab Report Chem 223 Spring Term 2021

Stella Kampmann

Abstract: the purpose of this experiment was to find the molecular weight of an unknown acid using its molecular weight. A diprotic acid with a known mass was titrated with a solution of NaOH with a known concentration. The molecular weight of the diprotic acid was found to be 115.98 which corresponded to the molecular weight of Maleic Acid with an error of 0.018%.

Introduction: the purpose of this experiment was to determine the weight of an unknown diprotic acid by finding its molecular weight through titration. An acid- base titration was set up using a known mass of an unknown acid and a solution of sodium hydroxide (NaOH) with a known concentration. The unknown acid was added to a beaker and the NaOH was carefully titrated in until the reaction had completed and a titration curve was generated. A diprotic acid is an acid with the general form H 2X that produces two H+ ions per acid molecule. Diprotic acids react with water in two stages: H2X (aq) ← → H+ (aq) + HX- (aq) HX- (aq) ← → H+ (aq) + X2- (aq) Because of the successive ionizations, the titration curves produced by diprotic acids can have two equivalence points as shown in figure 1. The equations for these acid-base reactions that occur between a diprotic acid and, in this case, sodium hydroxide are as follows: From the initial to the first equivalence point: H2X + NaOH ← → NaHX + H2O From the first equivalence point to the second: NaHX + NaOH ← → Na2X + H2O And lastly, from the beginning of the reaction through the second equivalence point:

H2X + 2NaOH ← → Na2X + 2H2O

Figure 1. General form of a titration curve for a diprotic acid showing that diprotic acids can have two equivalence points. At the second equivalence point all of the [H+] ions from the initial reaction have reacted with the present base (NaOH). At the second equivalence point, all [H+] ions from both reactions have reacted with the base so the volume of the base in the second reaction will be double that of the initial reaction. In this experiment a known mass of a diprotic acid is titrated with an NaOH solution with a known concentration. The molecular weight of the diprotic acid is found and reported in g/mol. The original sample of acid is then weighed, and its mass is found in grams. The moles are then found using the concentration of the NaOH titrant needed to reach the first equivalence point. The volume and concentration of NaOH are then used to obtain the moles of and NaOH. Using equation 3, the moles of the unknown acid are equal to the moles of NaOH at the first equivalence point. Once the grams of moles of the diprotic acid are known, the molecular weight

is calculated in g/mol. Either the first or second equivalence point can be used to calculate the molecular weight. Data:

Figure 2. Titration curve of sulfurous acid titrated to completion with NaOH showing two equivalence points. Concentration of base used: 0.1104 M Volume of H2SO4 used: 10 mL The initial pH of the H2SO4 used is 1.50. At the half equivalence point, the pH is 2, the first equivalence point is 4 and the second equivalence point is 10.

Results:

Figure 3. Plot of the first derivative of the titration of an unknown acid with NaOH.

Figure 4. Plot of the second derivative of the titration of an unknown acid with NaOH.

1 1st EP 2 2nd EP Figure 5. Plot of the pH change as the titration went to completion. Showing the ½ equivalence point, the 1st equivalence point, the 2nd ½ equivalence point, and the 2nd equivalence point.

Table 1. Showing the data gathered from the titration of the unknown acid and NaOH with the results. Moles of acid

Molar mass of acid

Identity of the acid

% error

Ka1 and Ka2

Equivalence point #1

0.00388 mol

115.98

Maleic Acid

0.018%

1.2 x 10-2

Equivalence point #2

0.00388 mol

115.98

Maleic Acid

0.018%

5.8 x 10-7

Discussion: the purpose of this experiment was to identify an unknown acid through titration. An unknown acid was titrated with NaOH and the resulting molecular weight and Ka values were used to decipher it. At the first equivalence point, all of the H+ ions react with NaOH from the first reaction. At the second equivalence point, all of the H+ ions from both reactions are reacted. This means that the

second equivalence point contains roughly twice the volume of NaOH as the first. Looking at Figure 5., one can see that the second equivalence point is the more clearly defined jump in pH. Diprotic acids that are titrated with a base such as NaOH will provide multiple equivalence points because they lose their protons one after the other. The 2:1 stoichiometric ratio between the acid and the NaOH will result in there always being twice as many moles of NaOH as there is acid. This allowed the grams of the unknown diprotic acid to be known which was then used to determine its identity. The experiment revealed that the unknown diprotic acid is Maleic acid (H2C4H2O4). The accepted weight of Maleic acid is 116 g and the experimentally determined weight was found to be 115.98. This gave a 0.018% error in calculation which allows a great deal of confidence in the determined value. The results of this experiment are solid and acceptable however there may have been some error in the experiment. To determine the equivalence point, the change in color of the solution is observed. While this can be useful, there is a chance that these results were skewed due to a lag time or a very quick change. This error would result in a skewed experimentally determined pH and the percent error would have been much higher. Performing more titrations would provide more accurate results however this experiment was a success nonetheless. Given the very low percent error of 0.018%, one can be very confident in the results of this experiment. The accepted Ka value for the first equivalence point matches the experimentally determined value however the accepted Ka value for the second equivalence point is slightly skewed with an accepted value of 5.4 x 10-7 and the determined value of 5.8 x 10-7. The molecular weight is where the most confidence in this experiment comes from.

Upon looking up the Ka values for other polyprotic acids, it can be said with confidence that Maleic Acid is the identity of the unknown acid. Calculations: Titrant = 0.155M Mass of Unknown Acid = 0.450 g Equ. Pt. 1 = 25mL NaOH pH = 4.13 Equ. Pt. 2 = 50 mL NaOH pH = 10.09 Mol NaOH at Equ. Pt. 1 and 2 1. Volume = 0.025 L 0.155 M = X mol / 0.025 L X mol = 0.155 M * 0.025 L = 0.0288 mol NaOH 2. Volume = 0.050 L 0.155 M = x mol / 0.050 L X mol = 0.155 M * 0.050 L = 0.0775 mol Mol H2X used 1. H2X + NaOH ← → NaHX + H2O 1:1 0.00388 mol HsX at Equ. Pt. 1 2. H2X + 2NaOH ← → NaHX + 2H2O 1:2 0.00388 mol H2X at Equ. Pt. 2 0.450 g / 0.00388 mol = 115.98 Maleic Acid ((116-115.98) / 115.98) * 100 = 0.018 Ka1 ½ * 25 = 12.5 10-1.89 = 1.2 x 10-2 Ka2 ½ * 75 = 37.5 10-6.2 = 5.8 x 10-7...