Mr. John Longo

Introduction

Acids are substances that donate hydrogen ions and bases are substances that accept hydrogen ions. Acids and bases react with each other by transferring hydrogen ions. One way to distinguish an acid is by its equivalent mass, which is the number of grams of the acid needed to transfer one mole of hydrogen ion to a base. For a monoprotic acid, which only transfers one hydrogen ion, its equivalent mass equals its molar mass. For a diprotic acid, which transfers two hydrogen ions, its equivalent mass equals half its molar mass. The equivalent mass of a base is simply the number of grams required to accept one mole of hydrogen ion. The equivalent mass of an acid or base is also equal to the mass of the acid or base titrated divided by the number of equivalents of the acid or base.

Acid strength is measured by how much it dissociates. This is determined by the amount of hydronium ion formed in water. Most often, it is expressed by pH, which equals the –log[H₃O⁺]. The equilibrium constant for an acid is represented by K_a = [H₃O⁺] [A^–] / [HA], where HA is the acid and A^– is the dissociated acid. When comparing K_a’s, most commonly pK_a is used, which is equal to –log K_a.

In this experiment, the pK_a of an unknown acid is determined by titrating it with NaOH and graphing its pH levels versus volume of NaOH titrated. The inflection point found by graphing is the equivalence point, and at half that volume is the half-equivalence point. At half equivalence, the [A^–] = [HA], so they cancel out in the equation K_a = [H₃O⁺] [A^–] / [HA], leaving K_a = [H₃O⁺]. With pH being known, K_a is found by [H₃O⁺] equaling 10^-pH, and pK_a = pH.

Experimental

First, about 2 g of an unknown weak acid was obtained. Then about 0.10 to 0.12 g of the acid was weighed out on an analytical scale. This mass was recorded to 4 decimal places. The acid was then transferred to a clean 125 mL Erlenmeyer flask. Distilled water was added to dissolve the acid, and then 3 drops of phenolphthalein indicator was added to the solution. Next, a buret was filled with NaOH solution and its initial reading was recorded to 2 decimal places. The acid solution was titrated with the NaOH solution until the acid solution turned and stayed pink. This final reading was recorded to 2 decimal places and the volume of NaOH used was calculated.

Another sample of unknown acid was weighed out using an analytical scale requiring about 25 mL of standardized NaOH solution. This mass was recorded to 4 decimal places. The acid was transferred into a clean 150 mL beaker, and was again dissolved using distilled water. A pH meter and electrode was then obtained. The electrode was rinsed with distilled water and put into the acid solution. Standardized NaOH solution was again used to titrate the acid. The initial buret reading was recorded to 2 decimal places and NaOH was added 0.5 mL at a time. After every portion added, the buret reading and pH meter reading were recorded to 2 decimal places. This was done until the pH spiked and stopped increasing significantly.

Results

Calculations

In order to find the volume of NaOH used, I subtracted the initial buret reading from the final reading (14.37 mL – 0.75 mL = 13.62 mL). To find the mass of unknown acid needed using 25 mL of NaOH solution, I used a proportion. The proportion was 0.1167 g / 13.62 mL = x / 25.00 mL, with x equaling the mass of unknown acid needed. To find the half-equivalence point, I divided the equivalence point volume by 2. I then just looked at the graph to find the pH at the half-equivalence point. The pK_a is equal to the pH, which was 4.62. The K_a is equal to the concentration of [H₃O⁺]. [H₃O⁺] is equal to 10^-pH, so using my numbers, 10^-4.62 = 2.40 x 10^-5. To find percent error, I took the theoretical pK_a – my found pK_a, divided by the theoretical pK_a­, and multiplied by 100%. Using my numbers, (4.69 – 4.62) / 4.69 x 100% = 1.49%. In order to find the number of equivalents of NaOH, I took the volume at the equivalence point (25.75 x 10^-3 L) and multiplied by the molarity (1.00 x 10^-1 M). The number of equivalents of acid titrated was equal to the equivalents of NaOH. Finally, to find the equivalent mass of the acid, I divided the mass of the acid used by the number of equivalents of acid. Using my numbers, 0.2152 g / 2.58 x 10^-3 equiv. = 83.41 g/equiv.

Discussion/Conclusions

Looking at the table in the lab book, the acid with the closest pK_a to my calculated pK_a is trans-crotonic. It has a pK_a of 4.69 while my calculated pK_a was 4.62. I feel that these numbers are very close, as the percent error is only 1.49%. The percent error for my calculated equivalent mass of the acid is also small at 3.11%. Error could have come from reading the graph, as I had to estimate the equivalence point. I should have taken measurements at smaller intervals when approaching the inflection point. This would have allowed me to read the graph more accurately. Error could have also come from the pH meter, which may not have been calibrated precisely. Lastly, error could have resulted if I inaccurately read the buret while recording the readings.

Introduction

A chemical reaction usually starts with reactants which react to yield products. Many times the reactants are completely used up to make products. However, the reactants sometimes do not completely turn into products. There is an equilibrium between the concentration of reactants and products. At equilibrium, the reactants turn into product and the products decompose into reactants at the same rate. This ratio of the products to reactants at equilibrium is represented by the equilibrium constant, or K. K is found by taking the concentration and order of the products and dividing by the concentration and order of the reactants. In this experiment, iron(III) ion reacts with thiocyanate ion to produce thiocyanatoiron (III). The reaction is represented by the following equation: Fe³⁺ + SCN^– <—-> FeSCN²⁺. Using a spectrophotometer, the absorbance of FeSCN²⁺ is measured at different concentrations. The absorbance in then put into Beer-Lambert’s law, A = εbc, to find concentration and ultimately the equilibrium constant.

Experimental

First, a clean cuvette was obtained, rinsed, and filled three-fourths full with 0.5 M HNO₃ solution. This was used as a the blank solution for the spectrophotometer, which was set at 447 nm. Next, a 100 mL volumetric flask was obtained and rinsed with distilled water. About 20 mL of 2.00 x 10^-3 M KSCN solution was then dispensed into a clean, dry 50 mL beaker. 10.00 mL of this solution was pipeted into the volumetric flask. The volumetric flask was then filled with distilled water to the line on the neck. The cap was put onto the volumetric flask and it was agitated to ensure consistency of the solution. This solution was then transferred into a clean, dry 250 mL beaker.

Next, about 20 mL of 1.00 x 10^-1 M Fe(NO­₃)₃ solution was dispensed into a clean, dry 50 mL beaker. 1.00 mL of this solution was pipeted into the 250 mL beaker with the KSCN solution. The solution was then mixed with a glass stirring rod. A second clean, dry cuvette was filled three-fourths full with this solution using a disposable Pasteur pipet. The spectrophotometer was blanked with the cuvette filled with 0.5 M HNO­₃ solution and the absorbance of the second cuvette was then measured and recorded.

The KSCN and Fe(NO­₃)₃ solution was then poured back into the 250 mL beaker. Another 1.00 mL of Fe(NO­₃)₃ solution was the pipeted into the 250 mL beaker. The cuvette used for measuring absorbance was filled with this solution using the disposable Pasteur pipet and was rinsed twice. It was finally filled three-fourth full with the solution in the 250 mL beaker. The spectrophotometer was blanked again with the 0.5 M HNO₃ solution and the absorbance of the solution in the other cuvette was recorded again. This process was repeated until 10.00 mL of Fe(NO­₃)₃ solution had been added to the 250 mL beaker.

Results

 

 

Average K: 169.13

Calculations

To find [Fe^*], I took the volume of Fe(NO₃)₃ used in liters, multiplied by the molarity of the Fe(NO₃)₃ concentration, 1.00 x 10^-1 M, and divided by the total volume of the solution. For example using the first mixture, (1.00 x 10^-3 L) (1.00 x 10^-1 M) / (0.101 L) = 9.90 x 10^-4 M. I did the same thing to find [SCN^*]. For example using the mixture 1, (10.00 x 10^-3 L) (2.00 x 10^-3 M) / (0.101 L) = 1.98 x 10^-4 M. To find [Fe(SCN)²⁺], I rearranged Beer’s law to c = A / (εb), with b equaling 1 cm and ε equaling 4700 L mol^-1cm^-1. Using the first mixture as an example,

c = 0.092 / (4700 L mol^-1cm^-1 x 1 cm), c = 1.96 x 10^-5 M. To find [Fe³⁺], I used the equation [Fe³⁺] = [Fe^*] – [Fe(SCN)²⁺]. For example, from mixture 1, [Fe³⁺] = 9.90 x 10^-4 M – 1.96 x 10^-5 M = 9.70 x 10^-4 M. To find [SCN-], I used the equation [SCN-] = [SCN^*] – [Fe(SCN)²⁺]. Using the first mixture, [SCN-] = 1.98 x 10^-4 M – 1.96 x 10^-5 M = 1.78 x 10^-4 M. To find K, I took the concentration and order of the products and divided by the concentration and order of the reactants. Using the concentrations from mixture 1, K = [Fe(SCN)²⁺] / ([Fe³⁺] x [SCN-]), K = 1.96 x 10^-5 M / (9.70 x 10^-4 M x 1.78 x 10^-4 M), K = 113.52. Finally, to find the average K, I added the 10 K values found and divided by 10.

Discussion/Conclusions

My results do not seem very accurate. Even during the experiment, my partner and I agreed that our absorbance readings did not seem right. The readings jumped a significant amount between the first and second readings and the fourth and fifth readings. After the seventh reading, the absorbance readings seemed to be evening out, as the difference between readings was getting smaller. When reading some of the absorbances, we took a second reading because a few times the absorbance reading was smaller than the previous. This did not seem right; the absorbance readings should have been going up each reading. Error may be attributed to the fact that a few drops of solution escaped the 250 mL beaker during transfers between the cuvette. Error could have also resulted from the beakers not being totally clean, or the pipetment of solutions could have been inaccurate. This would have caused the volumes and concentrations to be different than they actually were.

Introduction

The kinetics of a chemical equation is determined by its rate. The rate is the speed at which the reactants form into products. The rate is dependent on the concentrations and the orders of the reactants. One way to find the order is by first measuring the concentration of the products as time passes. A spectrophotometer is one tool that can measure relative concentration if the reactants change color as the form products. Graphing absorbance versus time, ln(absorbance) versus time, and 1/absorbance versus time will determine the order depending on which graph produces a straight line. In this experiment, crystal violet and NaOH form a complex that changes from transparent blue to colorless over time. The absorbance is measured using a spectrophotometer, and the rate law is then determined using this information.

Experimental

First, a spectrophotometer was turned on and set at a wavelength of 595 nm. Next, a cuvet was obtained, rinsed, and filled with deionized water. The outside of the cuvet was cleaned Kimwipe to get rid of smudges. The cuvet was then inserted into the spectrophotometer and the spectrophotometer was zeroed. Next, 10.0 mL of 0.010 M NaOH solution was dispensed into one clean 25 mL graduated cylinder and 10.0 mL of 1.50 x 10^-5 M crystal violet solution was dispensed into another clean 25 mL graduated cylinder. The solutions were then simultaneously poured into a clean 50 mL beaker. This mixture was mixed with a glass stirring rod for a few moments to ensure consistency. The cuvet was then rinsed with the mixture two or three times and was then filled with the mixture. The cuvet again cleaned with a Kimwipe and was inserted into the spectrophotometer. The absorbance reading was measured every minute for twenty minutes, starting when the cuvet was first put in. This process was then repeated, replacing the 0.010 M NaOH solution with 0.020 M NaOH solution.

Results

0.010 M NaOH Solution:

0.020 M NaOH Solution:

Discussion/Conclusions

For the graphs using 0.010 M NaOH, the plot of Absorbance vs. Time had the straightest line (R² = 0.9985), but the plot of lnA vs. Time also had a very straight line (R² = 0.9971). The plot of 1/A had the least straight line with R² equaling 0.9906. For the graphs using 0.020 M NaOH, lnA vs. Time had the straightest line (R² = 0.9998). Absorbance vs. Time and 1/A vs. Time were not nearly as straight, with their R² equaling 0.9646 and 0.9695, respectively. I think it is fairly safe to say that the rate equation is first order because lnA vs. Time overall yielded the straightest line in the two runs. The rate equation is rate = k [NaOH].

Introduction

The ideal gas law is used to define how gasses typically act. It is not precise, as gasses do not usually act ideally, but it works for most laboratory conditions. The ideal gas law is defined by PV = nRT (pressure in atm * volume in L = moles * constant * temperature in K). There is also a rule that states the total pressure of a system is equal to the sum of its partial pressures. Using this information, a small piece of metal magnesium is reacted with hydrochloric acid. The reaction is represented by the equation Mg (s) + 2HCl (aq) –> MgCl₂ (aq) + H₂ (g). Using the sum of partial pressures rule and the ideal gas law to find moles, the molar mass of Mg (s) can be calculated only knowing the pressure of H₂ (g).

Experimental

First, a strip of magnesium metal was obtained and weighed on an analytical balance. This weight was recorded to three significant figures. Next, the magnesium strip was wrapped around the tip of a copper wire, and was then encaged by the same copper wire by wrapping the wire around the magnesium. Following this, the tip of a buret was calibrated and this volume was recorded. Next, about 3 mL of concentrated HCl was put in the empty buret with the stopcock closed, and distilled water was then added until the buret was almost full. The copper wire with the magnesium was then lowered into the buret and held in place with a one-holed rubber stopper to make sure the wire would stay in place and would not fall to the bottom. Distilled water was squirted into the one-holed rubber stopped to make sure the buret was completely filled. The buret was then quickly inverted and clamped into a 600 mL beaker half filled with tap water. The water level on the buret was recorded after production of gas ceased. Finally, the distance between the height of water in the beaker and water in the buret was also recorded. The process was repeated for one more trial.

Results

Average percent error: 0.4525

Calculations

In order to find the volume of hydrogen, I took the total volume of the buret (50 mL plus the calibrated part) and subtracted my final reading. Using the results from my first trial, 50.0 mL + 1.20 mL – 9.40 mL = 41.80 mL. To find the pressure difference of water levels, I took the distance in difference of water levels, multiplied by the density of water, and divided by the density of Hg. From the first trial, 135 mm (1.0 g/mL) / 13.5 g/mL = 10.0 mm. The aqueous vapor pressure was found using a chart. At 23 ºC, the aqueous vapor pressure is 21.2 torr (or 21.1 mm Hg). Using the equation P_atm = P_{H2O (l)} + P_{H20 (g)} + P_{H2 (g)}, I could solve for the pressure of H₂ (g). Subbing in the numbers from the first trial, 762.76 torr = 21.2 torr + 10.0 torr + P_{H2 (g)}. P_{H2 (g)} = 731.6 torr. To find atmospheres, I multiplied by 1 atm/760 torr. 731.6 torr (1 atm/760 torr) = 0.9626 atm. I then rearranged the ideal gas law to find moles of hydrogen produced. n = PV/RT, n = 0.9626 atm * 0.04180 L / (0.0821 L atm mole^-1 K^-1 * 296 K), n = 0.001656 moles. In the equation for the reaction, there are an equal number of moles of hydrogen and magnesium used. To find the molar mass of magnesium, I took the mass of the magnesium used divided by the number of moles used. From trial 1, 0.0406 g / 0.001656 moles = 24.52 g/mol. To find percent error, I took the absolute value of the true value minus my calculated value divided by the true value and multiplied by 100%. Using trial 1, ( | 24.31 – 24.52 | ) / 24.31 * 100% = 0.8638%. Finally, to find the average percent error, I added the two percents and divided by two: (0.8638% + 0.04113%) / 2 = 0.4525%.

Discussion/Conclusions

My results are very good. An average percent error of 0.4525% is extremely small, so I feel that I performed the experiment well and had some good luck. I felt that when performing the experiment, I could have gotten some error from my temperature readings. The temperature seemed to fluctuate a couple degrees during the experiment and I thought that would negatively affect my final results. I also thought I could have gotten error from my difference in water levels measurement because it was hard to hold the ruler steady and read the ruler at the same time. Lastly, since the ideal gas law is not precise, error could have come from that. Other than those questionable factors, the experiment went smoothly.

Introduction

Titrations are often performed with acid and base solutions in order to determine their molarity. Once the enough acid is titrated into the base solution, the mixture will become colorless, indicating the mixture has become neutral. Depending on the chemical reaction at hand, there may be a certain ratio of moles of the acid to the base needed for neutrality. The same process can be formed with a redox reaction. IO₃^– (aq) + 6S₂O₃^-2 (aq) + 6H⁺ (aq) —> I^– (aq) + 3S₄O₆^-2 (aq) + 3H₂O (l) and HOCl (aq) + 2S₂O₃^-2 (aq) + H⁺ (aq) —> Cl^– (aq) + S₄O₆^-2 (aq) + H₂O (l) are two general chemical reactions which both can be used in a oxidation-reduction titration. Using the first equation, an unknown molarity of the S₂O₃^-2 component could be calculated and then used in the second equation to find the molarity of an unknown OCl^– component. In this experiment, the aforementioned process was performed in order to find the percent composition of household bleach.

Experimental

First, a 0.05 M Na₂S₂O₃ solution was made by dispensing 12.5 mL of a 1.0 M Na₂S₂O₃ into a 250 mL volumetric flask and then filling the volumetric flask with distilled water to the mark on the neck. The solution was then agitated to ensure consistency. Next 50 mL of a 1.507 * 10^-2 M KIO₃ stock solution was measured into a clean 100 mL beaker. Using a volumetric pipet, 10 mL of the KIO₃ solution was dispensed into a 300 mL Erlenmeyer flask. After that, about 25 mL of distilled water and 3 mL of a 3 M KI stock solution were also added and swirled in the Erlenmeyer flask. Finally, 2 mL of a 3 M H₂SO₄ stock solution was added to the flask, which changed the solution in the flask from being colorless to a deep brown.

Next, the 0.05 M Na₂S₂O₃ solution was put into a buret and was used to titrate the KIO₃/KI/H₂SO₄ mixture. The initial reading on the buret was recorded and titrating began. Once the mixture was a light yellow color, 1.0 mL of starch indicator was dispensed into the Erlenmeyer flask. This changed the mixture to a dark blue-black color. Titrating continued until the mixture became colorless. The final buret reading was recorded. This titrating process was repeated two more times.

For the next part of the experiment, an empty beaker was first weighed and recorded, then about 0.5 mL of bleach was dispensed into it and the new mass of the beaker was recorded. The bleach was then poured into a clean 300 mL Erlenmeyer flask. The beaker was rinsed with distilled water to ensure all the bleach transferred to the Erlenmeyer flask. Then about 25 mL of distilled water was added to the flask along with 3 mL of 3 M KI stock solution and 2 mL of 3 M H₂SO₄ stock solution. The mixture was swirled to obtain consistency, and then finally 5 drops of 3% ammonium molybdate catalyst was added to the mixture. The mixture was then titrated with the same Na₂S₂O₃ solution from earlier in the same manner for three trials.

Results

Standardization of 0.05 M Na₂S₂O₃ Solution:

Average Molarity, M: 4.902 * 10­­­^-2

Precision, ppt: 14.08

Determination of the Oxidizing Capacity of an Unknown Liquid Bleach:

Average Percentage NaOCl: 6.176 %

Precision, ppt: 22.67

Calculations

To find the number of moles of KIO₃ used, I used the equation Molarity = Moles/Liter. This equation can be rearranged to Moles = Molarity * Liter. Using my data, I subbed in 1.507 * 10­­­^-2 M * 0.01000 L = 1.507 * 10^-4 moles of KIO₃ used. To find the volume of Na₂S₂O₃ used for titration, I subtracted the initial buret reading from the final buret reading. For example, I did 19.91 mL – 1.30 mL = 18.61 mL for my first trial. To find the number of moles of Na₂S₂O₃ added, I looked at the chemical equation and saw that for every mole of KIO₃ used, 6 moles of Na₂S₂O₃ were used, so I multiplied the amount of KIO₃ used by 6 to find the number of moles of Na₂S₂O₃ used (6 * 1.507 * 10^{-4 =} 9.042 * 10^-4). To find the molarity of the Na₂S₂O₃, I used the equation Molarity = Moles/Liter. Using my data, I subbed in 9.042 * 10^-4 moles / 0.01861 L = 4.859 * 10­­­^-2 M. To find the average molarity, I added the values for the three trials and then divided by three. Then to find precision, I took the highest molarity (4.928 * 10­­­^-2 M) minus the lowest molarity (4.859 * 10­­­^-2 M), divided by the average molarity (4.902 * 10­­­^-2 M), and finally multiplied by 1000.

For the second part of the experiment, I did the same calculations for finding the volume of Na₂S₂O₃, the average, and precision as I did for the first part. To find the mass of the bleach, I subtracted the weight of the empty beaker from the weight of the beaker with the bleach in it. For the first trial, 27.37 g – 26.53 g = 0.84 g. To find the moles of Na₂S₂O₃, I again used the equation Moles = Molarity * Liters. In my first trial, 4.902 * 10­­­^-2 M * 0.02819 L = 1.382 * 10^-3 moles. To find the moles of NaOCl, I looked at the chemical equation and saw for every mole of NaOCl, there were two moles of Na₂S₂O₃, so I simply divided the number of moles of Na₂S₂O₃ by two (for the first trial, 1.382 * 10^-3 / 2 = 6.910 * 10^-4). In order to find the mass of NaOCl, I multiplied the number of moles of NaOCl by the molar mass of NaOCl (74.4419 g). Finally, to find the percent of NaOCl, I divided the mass of NaOCl by the mass of the bleach.

Discussion/Conclusions

My results seem very plausible. I know that standard bleach is diluted quite a bit, so 6.176 % of actual bleach could be very close to the correct answer. The lab procedures ran smoothly for the most part. I had to make a new mixture of KI, H₂SO₄, and bleach for the second part of the experiment when I put the wrong amount of H₂SO₄ in the Erlenmeyer flask, but that was not a big deal. When I obtained my bleach the first time, I think the dispenser was set at the wrong amount, as my weight of the bleach was significantly higher for my first trial. This did not seem to effect my final results however, as the percentage of NaOCl for that trial was almost exactly the same as trials two and three. Lastly, I think I probably had some error during my titrations. I think I did not stop the buret of Na₂S₂O₃ at the right moment. I am fairly sure a few extra drops came out, which would have effected my results, making the molarity of the Na₂S₂O₃ lower.