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Lab Report 4

by: Isabell Notetaker

Lab Report 4

Isabell Notetaker
DMACC - Ankeny Campus
GPA 3.52

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About this Document

This is a lab report on an acid based titration.
General Chemistry 2
Class Notes
25 ?




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This 3 page Class Notes was uploaded by Isabell Notetaker on Sunday September 25, 2016. The Class Notes belongs to at DMACC - Ankeny Campus taught by in Spring 2016. Since its upload, it has received 4 views. For similar materials see General Chemistry 2 in Science at DMACC - Ankeny Campus.


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Date Created: 09/25/16
Isabell Akers Lab 1/21/16 Lab Report - Acid Base Titration Introduction: The purpose of this lab was to find the molarity of oxalic acid (COOH) using the known molarity of the base solution, sodium hydroxide (NaOH) using titration. Materials & Methods:  Buret  Dropper  Ring stand  Scale (g)  Clamp to hold buret  Scupula  Flasks  Measuring boat  Beakers  Sodium Hydroxide (Base)  Glass stirring rod  Pheuolphetaleir (Indicator)  Graduated cylinders  Oxalic Acid (Acid)   Prepare Base (Sodium Hydroxide): 1. Weigh 1.999 grams of NaOH Combine NaOH with 250 mL of water to make NaOH solution.  Prepare Acid (Oxalic Acid): 1. Measure out 250 mL of water and combine with pre-measured COOH to make Oxalic Acid.  Procedure for Trials: 1. Put some of base in buret, record level of solution 2. Measure out 25 mL of acid into a flask 3. Add 2-3 drops of indicator to flask 4. Add base slowly to flask, a temporary pink color will appear, swirl until mixed and clear 5. When pick color is faint and permanent, we stopped adding base to acid. 6. Measure the amount of base left in the buret and record level of solution. 7. This was repeated three times  Results & Calculations: 1. Find out how much NaOH to weigh out a. Na (22.99) + O (16.0) + H (1.008) = 39.998g   b. 0.2 molarity = mole = 0.5 mol NaOH  0.250 mL  c. 0.5 moles of NaOH X molar mass of NaOH 39.998g = 1.999 g of NaCl  2. Take level B of the base minus level A, for all three trials.   Point B  Point A  Difference  Trial 1  29 mL  12.5 mL  16.5 mL  Trial 2  26.4 mL  10.2 mL  16.2 mL  Trial 3  25.9 mL  9.6 mL  16.3 mL  3. Find average difference a. 16.5 + 16.2 +16.3 = 49 > 49/3 = 16.33 mL 4. Moles needed of NaOH a. Molarity = moles / volume (L) b. 0.2 M = mol / 0.0163 L c. Moles = 0.00326 moles of NaOH needed to neutralize acid 5. Make a molar ratio between NaOH and COOH a. COOH + 2NaOH -> COO Na + 2H O 2 b. COOH = (0.00163mol) NaOH = (0.00326mol) ratio is 1:2 c. Molarity = 0.00163 moles / 0.025 L d. Molarity = 0.0652 COOH 6. Standard Deviation a. Average 3 trials = 16.33 b. For each trial subtract mean and square root i. 16.5 – 16.33 = 0.17 > square root of 0.17 = 0.412 ii. 16.2 – 16.33 = -0.13 > square root of 0.13 = .361 iii. 16.3 - 16.33 = -0.03 > square root of -0.03 = 0.173 1. Mean of those square roots 2. 0.412 + 0.361 + 0.173 = .315 3. Square of .315 = 0.561 c. Answer = 0.561  Conclusion:  The titration was successful when we noticed the pinkish color when the base was added to the acid. Our first trial turned out hot pink because we added to much base, this skewed our results and may be the reason our standard deviation is so high. The second and third trials were much closer together which leads me to believe that is the accepted value. Errors in the lab include; mistakenly adding to many drops of the indicator to acid, adding to much base to the acid, making a misjudgment in reading how much base was used, and not calculating the standard deviation correctly. Common lab error could be involved in all of these and could have affected the results.


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