Lab 10 - Titration Curves Lab Typed Report copy

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Chemistry

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Feb 20, 2024

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Lab 10 - Titration Curves  Report (must be typed) Graph Preparation. Using a ruler, draw a horizontal line at the equivalence point. At the intersection of this line and your curve, draw a perpendicular line to the x axis. Record the volume. At half that volume, draw a perpendicular line from the x axis so that it intersects your curve. From this intersection, draw a horizontal line so that it intersects the y-axis. Record this pH, it represents your Ka/Kb. Attach your prepared graphs to the back of the worksheet. (10 pts) Titration of a weak base (NH 3 ) with a strong acid (HCl) Fill-in the table below. (10 points) Show your computations clearly in your lab notebook. (4 pts) Value Initial volume of HCl (initial buret reading), mL 6.36 mL Final volume of HCl (final buret reading), mL 38.57 mL Drops, dp 562 dp Volume per drops, mL/dp 0.057 mL/dp Drops at equivalence point 428 Drops at half-equivalence point 214 pH at half-equivalence point 9.35 pOH at half-equivalence point 4.65 Experimental pK b of base 4.65 Percent error 2.16% Write an equation for the reaction that is taking place when the weak base is titrated with the strong acid. (3 pts) NH 3 (aq) + HCl (aq) HNH 4 (aq) + Cl (aq) 1
Circle the major species present in the beaker (excluding water) at various points in the titration curve of ammonia with hydrochloric acid. (8 pts) Write the dominant equilibrium reaction present at each of the following points in your titration. Each equilibrium should contain the major species you circled in the previous question. (3 pts each) a) Before the titration was started. NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) b) After 10.0mL of titrant were added. NH 4 + (aq) + HCl - (aq)  NH 4 Cl (aq) + H + (aq) c) At the equivalence point. NH 4 Cl (aq) + H + (aq) + OH - (aq)  NH 4 Cl (aq) + H 2 O (l) Why is the value of pH at the equivalence point different from neutral pH? Justify your answer using an appropriate chemical reaction. (3 pts) pH NH 3 HN 4 + H 3 O + OH pH = 7 Volume of HCl added. NH 3 HN 4 + H 3 O + OH NH 3 HN 4 + H 3 O + OH NH 3 HN 4 + H 3 O + OH 2
At the equivalence point, only NH 4 + and Cl - (NH 4 Cl) ions are present. NH 4 + is a conjugate acid of an weak base, NH 3 . So NH 4 + is relatively strong acid, and thus it will react with H 2 O to produce H 3 O + , making the solution acidic. Titration of a weak acid (CH 3 CO 2 H) with a strong base (NaOH) Fill-in the table below. (10 points) Show your computations clearly in your lab notebook. (4 pts) Value Initial volume of NaOH (initial buret reading), mL 1.95 mL Final volume of NaOH (final buret reading), mL 59.1 mL Drops, dp 172 Volume per drops, mL/dp 0.032 mL/dp Drops at equivalence point 1657 Drops at half-equivalence point 8.825 pH at half-equivalence point 4.475 pOH at half-equivalence point 9.524 Experimental pK a of acid 4.48 Percent error 6.04% Write an equation for the reaction that is taking place when the weak acid is titrated with the strong base. (3 pts) CH 3 COOH (aq) + NaOH (aq) → CH 3 COOHNa (aq) + H 2 O (l) Circle the major species present in the beaker (excluding water) at various points in the titration curve of acetic acid with sodium hydroxide. (8 pts) pH pH = 7 HC 2 H 3 O 2 HC 2 H 3 O 2 H 3 O + OH HC 2 H 3 O 2 HC 2 H 3 O 2 H 3 O + OH HC 2 H 3 O 2 HC 2 H 3 O 2 H 3 O + OH HC 2 H 3 O 2 HC 2 H 3 O 2 H 3 O + OH 3
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Write the dominant equilibrium reaction present at each of the following points in your titration. Each equilibrium should contain the major species you circled in the previous question. (3 pts each) a) Before the titration was started. CH 3 COOH (aq) + H 2 O (l) ↔ CH 3 COO - + H 3 O + (l) b) After 10.0mL of titrant were added. CH 3 COOH (aq) + OH - (aq) ↔ CH3COO - (aq) + H 2 O (l) c) At the equivalence point. CH 3 COOH (aq) + OH - (aq) ↔ CH3COO - (aq) + H 2 O (l) Why is the value of pH at the equivalence point different from neutral pH? Justify your answer using an appropriate chemical reaction. (3 pts) The pH at the equivalence point is not equal to 7 when the salt formed in the neutralization is either acidic or alkaline instead of neutral salt. When a weak acid is neutralized with a strong base, an alkaline salt is formed due to which the resulting solution is basic in nature and the pH value for the solution is greater than 7. This is true for the following reaction: CH 3 COOH+NaOH→CH 3 COONa+H 2 O. Since CH 3 COO⁻Na⁺ is a basic salt, the pH will be greater than 7, and thus, basic in nature. Volume of HCl added. 4
Error Analysis Using your data, discuss the accuracy of your results in determining the pK a of acetic acid and the pK b of ammonia . Refer to the appropriate statistical analysis data in your discussion. (4 pts) In acquiring the pK b of ammonia, our percent error was 2.16%. Since this value lies in between one and five percent, the accuracy of the date collected was very good. However, in acquiring the pK a of acetic acid, we found the percent error to be 6.04%. Since this is between five and ten percent, the accuracy of our data is still good, but improvement is possible. 5