D5

docx

School

Athabasca University, Athabasca *

*We aren’t endorsed by this school

Course

218

Subject

Chemistry

Date

Jan 9, 2024

Type

docx

Pages

Uploaded by ElderHawk3269

1 November 14, 2023 Experiment D5: Titration Curves Purpose The purpose of this experiment is to develop two potentiometric titration curves and mathematically determine the equivalence point based on measurements of pH values of, (1) a solution of a strong acid, hydrochloric acid (HCl), with various volumes of a strong base, sodium hydroxide, (NaCl), HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l) (1) hydrochloric acid Sodium hydroxide Sodium Chloric and (2) a solution of a weak acid, acetic acid (CH 3 CO 2 H), with various volumes of a strong base, sodium hydroxide, (NaCl), CH 3 CO 2 H (aq) + NaOH (aq) CH 3 CO 2 - NA + (aq) + H 2 O (l) acetic acid Sodium hydroxide Sodium Acetate CH 3 CO 2 - Na + (aq) + H 2 O (l) CH 3 CO 2 H (aq) + NaOH (aq) In order to determine the unknown concentration of sodium hydroxide from titration curve (1) and subsequently determine the unknown concentration of acetic acid in titration curve (2). Procedure The experiment was preformed using the same methodology as described in Experiment D5 of the Chemistry 218 Laboratory Manual pp. 85-92 and experimental data and observations were provided by the Athabasca University Chemistry 218 Laboratory with no revisions. Results Table 1. Total volume (mL) of sodium hydroxide, NaOH, added to the strong acid, hydrochloric acid (HCl), and resulting pH measurement taken with a glass/calomel electrode. Total Volume of NaOH (mL) pH (strong base) (strong acid) (salt of a strong acid) (2) (strong base) (weak acid) (salt of a weak acid) (3)

2 0.00 1.41 0.50 1.42 1.00 1.45 1.50 1.48 2.00 1.51 2.50 1.54 3.00 1.57 3.50 1.60 4.00 1.64 4.50 1.68 5.00 1.73 5.50 1.78 6.00 1.84 6.50 1.90 7.00 1.98 7.50 2.09 8.00 2.19 8.50 2.33 9.00 2.55 9.20 2.65 9.40 2.86 9.60 3.17 9.80 4.16 10.00 6.17 10.20 7.50 10.40 9.81 10.60 10.41 10.80 10.68 11.00 10.94 11.50 11.28 12.00 11.48 12.50 11.61 13.00 11.71 13.50 11.79 14.00 11.85 14.50 11.90 15.00 11.95 15.50 11.99 16.00 12.03 16.50 12.06 17.00 12.09 17.50 12.11 18.00 12.14

3 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 16.00 18.00 20.00 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 f(x) = 0.86 x − 1.9 Total NaOH added (mL) pH Figure 1. Total volume (mL) of sodium hydroxide (NaOH) added to the strong acid, hydrochloric acid (HCl), and resulting pH measurement taken with a glass/calomel electrode. Table 2. Total volume (mL) of sodium hydroxide (NaOH) added to the weak acid, acetic acid (CH 3 CO 2 H), and resulting pH measurement taken with a glass/calomel electrode. Total Volume of NaOH (mL) pH 0.00 3.04 0.50 3.44 1.00 3.65 1.50 3.82 2.00 3.96 2.50 4.08 3.00 4.17 3.50 4.26 4.00 4.34 4.50 4.42 5.00 4.50 5.50 4.57 6.00 4.65 6.50 4.73 7.00 4.80 7.50 4.90 8.00 4.98 8.50 5.06 9.00 5.16 9.20 5.21

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

4 9.40 5.27 9.60 5.32 9.80 5.37 10.00 5.43 10.20 5.48 10.40 5.57 10.60 5.67 10.80 5.77 11.00 5.89 11.50 6.48 12.00 9.30 12.50 10.65 13.00 11.09 13.50 11.34 14.00 11.52 14.50 11.63 15.00 11.72 15.50 11.80 16.00 11.85 16.50 11.90 17.00 11.94 17.50 11.98 18.00 12.01 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 16.00 18.00 20.00 0.00 2.00 4.00 6.00 8.00 10.00 12.00 14.00 f(x) = 0.5 x + 2.48 Total NaOH added (mL) pH Figure 2. Total volume (mL) of sodium hydroxide (NaOH) added to the weak acid, acetic acid (CH 3 CO 2 H), and resulting pH measurement taken with a glass/calomel electrode.

5 Calculations 1. Line tangent to the upper portion of the pH titration curve, y = 0.857x + 1.424 Line tangent to the lower portion of the pH titration curve, y = 0.857x – 5.225 Half-way point, (1.424 + (-5.225))/2 = -1.9005 y = 0.857x – 1.9005 2. Equivalence point between sodium hydroxide and hydrochloric acid (10.18, 6.824) pH = 6.82 NaOH = 10.18 mL 3. Initial dilution of HCl [HCl] = 0.10 mol/L V HCl = 0.010 L V = 0.030 L [HCl] = (0.10 mol/L)(0.01 L)/(0.030 L) = 0.033 mol/L C 1 V 1 =C 2 V 2 [NaOH] = (0.033 mol/L) (0.030 L) / (0.01018 L) = 0.098 mol/L 4. Line tangent to the upper portion of the pH titration curve, y = 0.5x + 4.59 Line tangent to the lower portion of the pH titration curve, y = 0.5x + 0.37 Half-way point, (4.59 + 0.37)/2 = 2.48 y = 0.5x + 2.48 Equivalence point, (12.2, 8.58) pH = 8.58 NaOH = 12.2 mL 5. [CH 3 CO 2 H] = (0.098 mol/L) (0.0122 L) / 0.030 L = 0.04 mol/L [CH 3 CO 2 H] = (0.04 mol/L) (0.030 L) / (0.010 L) = 0.11 mol/L

6 Questions 1. Calculate the pH at the equivalence point in a titration between 10.00 mL of acetic acid (0.100 mol  L  1 ) and sodium hydroxide (0.100 mol  L  1 ). How does this value compare with the value you obtained in the experiment? Volume of NaOH V CH3CO2H = 0.010 mL [CH 3 CO 2 H] i = 0.100 mol/L [NaOH] i = 0.100 mol/L C 1 V 1 =C 2 V 2 V NaOH = (0.100 mol/L) (0.010 L) / (0.100 mol/L) = 0.010 L V tot = 0.010 + 0.010 = 0.020 L CH 3 CO 2 H = (0.100 mol/L) (0.010 L) = 1 x 10 3 mol NaOH = (0.100 mol/L) (0.010 L) = 1 x 10 3 mol [CH 3 CO 2 H] = 1 x 10 3 mol / 0.020 L = 0.05 mol/L [NaOH] = 1 x 10 3 mol / 0.020 L = 0.05 mol/L [CH 3 CO 2 H] [OH] [CH 3 CO 2 - ] [H 2 O] I 0.05 0.05 ≈ 0 - C - 0.05 - 0.05 + 0.05 - E 0 0 0.05 - [CH 3 CO 2 - ] [H 2 O] [CH 3 COOH] [OH - ] I 0.05 - 0 0 C - x - + x + x E 0.05 - x - x x k b = [ CH 3 COOH ] ¿¿ 5.3 x 10 − 10 = x 2 0.05 − x ← x is small approximation

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

7 x = √ ( 5.3 x 10 − 10 ) ( 0.05 ) = 5.15 x 10 -6 = [OH - ] pOH =− log ¿¿ ¿ − log ( ¿ 5.15 x 10 − 6 ) ¿ ¿ 5.29 pH = 14 – pOH = 14 – 5.29 = 8.71 This pH value differs from our observed equivalence point, pH of 8.58, in the reaction of acetic acid and sodium hydroxide by an increase of 0.13 by relatively equivalent. 2. In the table shown in the “Results and Calculations” section of this experiment, the pH of the solution before sodium hydroxide has been added is indicated as being 1.00. This corresponds to the pH of a 0.10 mol  L  1 solution of hydrochloric acid. Your initial pH reading was probably not 1.00. Why? The pH of the hydrochloric acid solution prior to the addition of sodium hydroxide in part A of the experiment was observed to be 1.41. This is due to the dilution of the 0.010 L 0.100 mol/L hydrochloric acid solution to 0.033 mol/L by the addition of 0.020 L of distilled water, lowering the value of the pH. 3. In a potentiometric titration between weak acid and a strong base, such as the one carried out in Part B of this experiment, the pKa of the weak acid is equal to the pH of the solution, when half of the volume of sodium hydroxide required to reach the equivalence point has been added. From your results, determine the pKa of acetic acid. Compare your value with the literature value. Literature value pKa = 4.76 Experimental value NaOH at equivalence point / 2 = 12.2 mL / 2 = 6.1 mL pKa = pH ≈ 4.68

8 The calculated experimental value of pKa of acetic acid in this experiment was approximately 4.68, 0.08 less than the literature value of 4.76. Conclusion A steep potentiometric titration curve of a strong acid, hydrochloric acid, and a strong base, sodium hydroxide, was constructed with an equivalence point of pH 6.82 and sodium hydroxide volume 10.18 mL. The concentration of sodium hydroxide was then determined from this titration curve to be 0.098 mol/L. A second flatter potentiometric titration curve was constructed of a weak acid, acetic acid and the same concentration of sodium hydroxide, with an equivalence point of pH of 8.58 and sodium hydroxide volume of 12.2 mL. From this the concentration of acetic acid was determined to be 0.11 mol/L. These data demonstrated the relationship between the strength of the acid used and the steepness of the titration curve produced.

Related Documents

Copy of Exp5_W18_Page1.docx

Chem Lab Report 1.docx

lab 5 enthalpy of ice.pdf

102 _ PREOPERATIVE CARE - CH 18.docx

CHM2132_DGD_assign7_2023 2.pdf

Chapter 03 .pdf

Chapter 9.pdf

Chapter 10 .pdf

Related Questions

A chemistry graduate student is given 125. mL of a 1.60 M dimethylamine ((CH,) NH) solution. Dimethylamine is a weak base with K; = 5.4 × 10¯*. What mass of (CH,) NH,Cl should the student dissolve in the (CH, NH solution to turn it into a buffer with pH = 10.46? You may assume that the volume of the solution doesn't change when the (CH, NH,Cl is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits.

A chemistry graduate student is given 500. mL of a 1.80M hydrocyanic acid (HCN) solution. Hydrocyanic acid is a weak acid with K,= 4.9 x 10 – 10 What mass of KCN should the student dissolve in the HCN solution to turn it into a buffer with pH =9.70? You may assume that the volume of the solution doesn't change when the KCN is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits. 国国回

Buffer capacity is a measure of a buffer solution's resistance to changes in pH as strong acid or base is added. Suppose that you have 145 mL of a buffer that is 0.300 M in both acetic acid (CH₂COOH) and its conjugate base (CH3COO-). Calculate the maximum volume of 0.170M HCl that can be added to the buffer before its buffering capacity is lost. volume: esc ! 1 ← 2 #3 с $ 4 G Search or type URL 27 20 % MacBook Pro 5 ^ 6 & 7 +00 8 ( 9 ) mL 0 1

GW 18b 1. A 20.0 mL sample of a 0.240 M hydrofluoric acid (HF) solution is titrated with 0.200 M NaOH. Determine: (a) pH of the acid solution before any base is added; (b) volume (in mL) of base needed to get to the equivalence point; (c) pH halfway to equivalence point; (d) pH at equivalence point; (e) pH when 0.100 mL NaOH is added beyond the equivalence point. (K₂ of HF = 7.1 × 104)

The hydrogen tartrate anion itself is a weak acid and slightly dissociates in aqueous solution. Htar (aq) H* (aq) + Tar²-(aq) However, HTar is a strong enough acid that it can be titrated with a strong base like sodium hydroxide and this neutralization reaction proceeds to completion. Thus we can analyze a solution for HTar by titration with a standard strong base solution such as aqueous NaOH. HTar (aq) + OH(aq) Tar²(aq) + H2O (a) If 50.0 mL of HTar in water solution was titrated by 16.0 mL of 0.098 M NaOH. Calculate the concentration of HTar", [HTar']? (b) Calculate Ksp in this case? Show your result with the proper number of significant figures. (Hint: use the above result and Ksp = [K*] [HTar']) Type your answer...

A buffer is made using 100.0 mL of 0.100 M CH3 CH2 COOH (propanoic acid) and 100.0 mL of 0.100 M NACH3 CH2 COO (sodium propanoate). A) Explain in your own words what will occur (at the molecular level) when an nitric acid is added to the buffer? What would be the effect on the pH? B) Explain in your own words what will occur when LIOH is added tot he buffer? What would be the effect on the [H+]?

consider the titration of 50.0 mL of 0.10 M acetic acid with NaOH. drag and drop each amount of NaOH added (to the acetic acid) Into the appropriate resulting pH. In other words, determine the pH of the final solution after each volume of NaOH has been added. Will the resulting solutions be acidic, basic, or neutral? Consider the stration of 50.0 ml of 0.10 M acetic acid (HC₂H₂O₂. K, -18 x 10) with NaOH. Drag and drop each amount of NaOH added to the acetic acid) into the appropriate resulting pH. In other words determine the pH of the final solution after each volume of NaOH has been added. Will the resulting solution be acidic, basic, or neutra? Acidic Neutral Basic Drag and drop your selection from the following list to complete the answer 25.0 mL (total) of 0.10 M NaOH has been added (the halfway point) 50.0 mL. (total) of 0.10 M NaOH has been added (the equivalence point) 10.0 mL (total) of 0.10 M NaOH has been added 60.0 mL (total) of 0.10 M NaOll has been added No NaOH has been…

SEE MORE QUESTIONS

Recommended textbooks for you

General Chemistry - Standalone book (MindTap Cour...

Chemistry

ISBN:9781305580343

Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell

Publisher:Cengage Learning