experiment-four-acid-base-titrations_compress (1)

pdf

School

University of Ottawa *

*We aren’t endorsed by this school

Course

1300

Subject

Chemistry

Date

Feb 20, 2024

Type

pdf

Pages

9

Uploaded by ProfessorEnergy10817

Report
Experiment four acid base titrations Principles of Chemistry (CHM 1311) Experiment 4: Acid-Base Titrations Chemistry 1311 – Section Z03 Rashmi Venkateswaran Stacey Welsford 300135964 TA: Vanessa Susevski University of Ottawa November 7 , 2019 th 0 0
Introduction Acids and bases are frequently met chemical substances that can be defined in several ways. An acid is a substance that behaves as a proton donor when dissolved in water, while bases are proton acceptors, or hydroxide ion donors in water (Acid Base Titrations 77). A proton is an H + ion while a hydroxide ion is OH (Acid Base Titrations 77). For an example, the acid HNO - 3 dissociates in water according to the following equation: HNO 3(aq) H + NO + (aq) 3 (aq) - Since there is one lone proton on the products side that is not attached to NO , it can be said that 3 - HNO 3 donated its proton and is therefore an acid. When dissolved in water, NaOH separates from its hydroxide ion: NaOH (aq) Na + OH + (aq) - (aq) Since NaOH donates its hydroxide ion to the product side of the reaction, it is shown that NaOH is a base. The concentration of acid or base in a solution can be determined using the equation c = n ÷ V, where c represents concentration, n represents moles and v represent volume of the solution in litres. The moles of solute can be determined by dividing its mass in grams by its molar mass in grams per mole: n = m ÷ MM. The acidity of a substance can be determined using the pH scale. The pH scale is a logarithmic scale where an increase or decrease of 1 results in changes by multiples of 10 for the concentration of H (Silberberg 732). A value of 7.0 on the pH scale resembles a neutral 3 O + (aq) solution, a value less than 7.0 would be classified as acidic, and a value more than 7.0 would be classified as basic. The more a pH value deviates from 7.0, the stronger the acidity or basicity of the substance will be. Acids and bases endure neutralization reactions where they form a salt and water, and the equivalence point refers to the point where the amount of acid added to a solution is enough to neutralize the amount of base present (Acid Base Titrations 78). The less an equivalence point deviates from an endpoint, the more accurate the equivalence point is (Bard 364). If a base was being titrated with an acid, the amount of acid required to titrate the base can be found by first determining the moles of the base present; this can be done by re-arranging the concentration equation and dividing the concentration of the base its the volume to isolate the amount of moles. Then, referring to a balanced chemical equation, the coefficients associated with the acid-base ratios must be used while calculating the amount of acid required. For example, if the coefficients of the balanced equation show a 1:2 ratio between the base and the acid, the moles of the base obtained from the previous equation must be multiplied by 0.5; in this case, this is because the moles of the acid is twice the amount of the moles of the base. The volume of acid required can then be determined by re-arranging the mole equation, c = n ÷ V, to V = n ÷ c. The 0 0
following equation can then be used to isolate the final required quantity, where b is the stoichiometric coefficient of the base and a is the coefficient of the acid (Acid Base Titrations 77): c base ×V base = b a c acid ×V acid For safety precautions in this laboratory, lab goggles and lab coats will be worn throughout the entire experiment as skin or eye contact with acidic or basic solutions has the potential result in harmful and permanent consequences. Procedure As outlined in Lab Manual (Rashmi Venkateswaran . “Oh how bitter a thing it is…” Acid Base Titrations. 2019. Pages 78-84) Data/Observations Standardizing NaOH – Trial One (Figure One and Table One) Number of Drops of Phenolphthalein Volume of HCl Acid (Initial) Volume of HCl Acid (Final) HCl Acid that Flowed Out Volume of NaOH Base (Initial) Volume of NaOH Base (Final) 3 11.51 mL 21.49 mL 9.98 mL 40.10 mL 30.11 mL Standardizing NaOH – Trial Two (Figure Two and Table Two) 0 0
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Number of Drops of Phenolphthalein Volume of HCl Acid (Initial) Volume of HCl Acid (Final) HCl Acid that Flowed Out Volume of NaOH Base (Initial) Volume of NaOH Base (Final) 3 7.49 mL 17.45 mL 9.96 mL 40.20 mL 32.10 mL Standardizing NaOH – Trial Three (Figure Three and Table Three) Number of Drops of Phenolphthalein Volume of HCl Acid (Initial) Volume of HCl Acid (Final) HCl Acid that Flowed Out Volume of NaOH Base (Initial) Volume of NaOH Base (Final) 3 13.51 mL 23.51 mL 10.00 mL 32.10 mL 23.50 mL Unknown Acid – Trial One (Figure Four and Table Four) Number of Drops of Volume of Unknown Volume of Unknown Unknown Acid that Volume of NaOH Base Volume of NaOH Base 0 0
Phenolphthalein Acid (Initial) Acid (Final) Flowed Out (Initial) (Final) 3 1.00 mL 11.05 mL 10.05 mL 41.90 mL 32.10 mL Unknown Acid – Trial Two (Figure Five and Table Five) Number of Drops of Phenolphthalein Volume of Unknown Acid (Initial) Volume of Unknown Acid (Final) Unknown Acid that Flowed Out Volume of NaOH Base (Initial) Volume of NaOH Base (Final) 3 11.05 mL 21.10 mL 10.05 mL 32.10 mL 25.90 mL Unknown Acid – Trial Three (Figure Six and Table Six) Number of Drops of Phenolphthalein Volume of Unknown Acid (Initial) Volume of Unknown Acid (Final) Unknown Acid that Flowed Out Volume of NaOH Base (Initial) Volume of NaOH Base (Final) 3 11.10 mL 21.10 mL 10.00 mL 25.90 mL 5.20 mL 0 0
Calculations 1. V 2 = 249 m L H 2 O + 4.61 m L NaOH V 2 = 253.61 mL c c 1 V 1 = 2 V 2 6 M × 4.61 mL = c 2 × 253.61 mL C 2 = 0.1 M 1. NaOH + HCl H O + NaCl 2 c base ×V base = b a c acid ×V acid c base × 2.316 mL = 1 1 1 ( 0.1 M ) acid × 9.98 mL c base = 0.998 M ∙mL 2.316 mL c base = 0.43 M Data for calculation was obtained from Table 1 and Figure 1. c base ×V base = b a c acid ×V acid c base × 7.593 mL = 1 1 1 ( 0.1 M ) acid × 9.96 mL c base = 0.996 M ∙mL 7.593 mL c base = 0.13 M Data for calculation was obtained ¿ Table 2 Figure 2. c base ×V base = b a c acid ×V acid c base × 1.508 mL = 1 1 1 ( 0.1 M ) acid × 10.00 mL c base = 1 M ∙mL 1.508 mL c base = 0.66 M Data for calculation was obtained from Table 3 and Figure 3. 2. H 2 A (aq) + NaOH NaHA (aq) (aq) + H 2 O (l) NaHA (aq) + NaOH Na + H (aq) 2 A (aq) 2 O (l) H A 2 A (aq) + 2NaOH Na (aq) 2 (aq) + 2H 2 O (l) c base ×V base = b a c acid ×V acid 0.06 M × 10.39 mL = 2 1 c acid × 10.05 mL 0 0
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
c acid = 0.6234 M ∙mL 20.10 mL c acid = 0.03 M Datafor previouscalculationwas obtained ¿ Table 4 Figure 4. c base ×V base = b a c acid ×V acid 0.06 M × 13.62 mL = 2 1 c acid × 10.05 mL c acid = 0.8172 M ∙mL 20.10 mL c acid = 0.04 M Data for calculation was obtained from Table 5 and Figure 5. c base ×V base = b a c acid ×V acid 0.06 M × 13.46 mL = 2 1 c acid × 10.00 mL c acid = 0.8076 M∙mL 20.00 mL c acid = 0.04 M Datafor previouscalculationwas obtained ¿ Table 6 Figure 6. Discussion Though this experiment was performed as precisely as possible, given the limited amount of time in the laboratory, there were sources of error that could not be adjusted. For example, the drop counter did not properly calibrate with Logger Pro at the beginning of the experiment. During trial 1 with the hydrochloric acid, the burette reading for volume of NaOH and the value given by LabQuest 2 differed by more than 10mL. Since the values were extremely inaccurate, the data displayed in Table 1 and Figure 1 is invalid. After the first trial, the drop counter was re- calibrated, and LabQuest 2 was restarted. However, when instructions were carefully followed to calibrate the drop counter for the second time, the amount of dilute NaOH that flowed out of the 0 0
plastic burette was not exactly the same as the value given on Logger Pro for each trial; typically, the values were 1-2mL higher on Logger Pro than the visual burette measurements. Since the Logger Pro volume value showed the relationship with pH of the solution, the Logger Pro values were recorded for the calculations and graphs. However, since Logger Pro recorded more volume than there truly was, this means that there was a source of error as the system readings suggest that more NaOH was needed to change the colour of the solution. Logger Pro also suggests that the solutions were more concentrated with NaOH than they truly were. Future experimenters should use a different LabQuest system if the calibration with the drop counter is not exact, or retry the calibration enough times in order to ensure that they have the most precise calibration possible. However, given that laboratory time was limited to 3 hours, the experiment had to continue with a slightly inaccurate calibration in order to reach completion in time. The volume of concentrated NaOH used in the beginning of the lab does not matter, as long as the volume is kept constant throughout the entire experiment. Decreasing the amount of concentrated NaOH while keeping the amount of water added the same would make the solution in the plastic burette more dilute; thus, it is likely that it would take more dilute NaOH to cause the colour change in the acidic solutions in the beaker. However, as long as the amount used at the beginning does not change throughout the experiment (i.e. the solution in the burette does not become more dilute or more concentrated due to changing H O and NaOH proportions), 2 then the process would still be scientifically correct. The concentration of NaOH was determined before being used as the basic solution is unstable; specifically, sodium hydroxide solutions are unstable because they have the tendency to absorb atmospheric carbon dioxide from the surroundings which changes the [NaOH] itself. During each trial for each acid, the colour change occurred at a lesser volume compared to the equivalence points determined by Logger Pro. This means that the observed equivalence point occurred before the more accurate Logger Pro equivalence point determination. This may be because as soon as a faint pink colour appeared in the solution, the volume was recorded. However, the pink colour became darker and more prominent with increased volume of dilute NaOH; it is likely that the equivalence point was recorded too soon. To resolve this source of error, the observed value for the change of colour should be recorded when the solution reaches its darkest appearance. Appendix Raw Data 0 0
Reference Bard, A. J., & Simonsen, S. H. (n.d.). The General Equation for the Equivalence Point Potential in Oxidation-Reduction Titrations . Austin. Retrieved November 5, 2017, from http://bard.cm.utexas.edu/resources/Bard-Reprint/8.pdf Rashmi Venkateswaran . “Oh how bitter a thing it is…” Acid Base Titrations. 2019. Pages 78-84 Silberberg, Martin S., et al. Chemistry: The Molecular Nature of Matter and Change . McGraw- Hill Education, 2018. 0 0
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help