Acid-Base Titrations

Basic Lab Report Rubric Student Name: ________________________________ Acid-Base Titrations Experiment Title: __________________________________________ Introduction ¥ ¥ Objectives stated Essential concepts complete ¥ ¥ Some objectives stated Essential concepts incomplete ¥ Procedure not submitted Procedure ¥ Signed procedure included Data and Results Scientific Integrity ¥ Neat and legible ¥ Errors are crossed out with one strike and rewritten neatly Significant Figures & Units ¥ Correct units are used ¥ Correct S.F. representing glassware and equipment used ¥ ¥ Penmanship can be improved Errors are scribbled ¥ ¥ Messy Use of pencil ¥ ¥ Few errors on units Few S.F. errors representing glassware and equipment used ¥ ¥ Many errors on units Many S.F. errors representing glassware and equipment used ¥ Lacking accuracy and/or precision Not all data shown ¥ ¥ Poor accuracy and/or precision Missing a significant amount of data ¥ ¥ ¥ ¥ ¥ Has proper title Axes labeled & units shown May be missing some data points Best fit curve or line present May be missing the equation of the line shown ¥ ¥ Not proper title Improper labeling of axes labeled and/or no units shown May be missing some data points Missing best fit curve or line Missing the equation of the line ¥ ¥ ¥ ¥ Numbered as per lab manual Has titles describing work Formula may be shown Some incorrect answers ¥ ¥ ¥ Data ¥ ¥ Accurate and precise All data shown ¥ Graphs (if applicable) ¥ ¥ ¥ ¥ ¥ Has proper title Axes labeled & units shown Data points are shown Best fit curve or line present Equation of the line shown ¥ ¥ Sample Calculations ¥ ¥ ¥ ¥ Numbered as per lab manual Has titles describing work Formula is shown Correct answers Not numbered as per lab manual Lack of titles describing work Formula may not be shown Questions (pre lab and/or post lab questions): Only randomly selected questions will be graded ¥ ¥ Selected question(s) is/are correctly answered All questions have been answered Selected question(s) is/are correctly answered Not all questions have been answered ¥ ¥ ¥ ¥ Sentence stating results Includes proper sources of error Discussion incomplete ¥ Sentence stating results Includes sources of error that are not correct (human error) Discussion complete ¥ ¥ Has some references (1-2) Improper formatting ¥ No references ¥ ¥ ¥ Selected question(s) is/are incorrect Not all questions have been answered Conclusion ¥ ¥ ¥ Sentence stating results Includes proper sources of error Discussion complete ¥ ¥ References ¥ ¥ 1. 2. 3. 4. 5. 6. 7. 8. Has required references (2-3) Proper format Every student has to submit their own lab report. At the end of every laboratory session, make sure your instructor signs your data sheet. Reports must be typed (calculations also). You will be graded based on significant figures and units Marks will be deducted if this format is not respected (follow presentation order from template) Each section must be on a separate page and have a proper title (same as rubric grid) Use this page as your title page and template for your lab report. Report must be in font times or times new roman and size 11 or 12. Acid-Base Titrations Introduction: Hypothesis: Objective: Material used: Sample Calculations Beaker B Moles of NaOH = mass NaOH / M.W NaOH = 0.18 g / 40.01 g/mol = 4.5 x 10^-3 mol NaOH (before and after dilution) Initial concentration [NaOH] = mol NaOH / V. solution = 0.0045 mol NaOH / 0.25 L = 0.018 M (250 mL ~ 0.25 L) Volume used: 35 mL + 15 mL = 50 mL Final concentration [NaOH]: C1¥V1 = C2 ¥ V2 [NaOH] concentration x V. concentration = [NaOH] dilution x V. dilution [NaOH] dilution: [NaOH] concentration x V. concentration / V. dilution Final concentration [NaOH]: 0.018 M x 0.035 L / 0.050n L = Beaker C Initial concentration [NaOH]: 1.3 x 10^-2 M Moles of NaOH: 0.013 M = x / 0.025 L (0.013) (0.025) = x x = 3.25 x 10^-4 mol NaOH Volume used: 25 mL + 25 mL = 50 mL Final concentration [NaOH]: 0.013 M x 0.025 L / 0.050 L = Beaker D Initial concentration [NaOH]: 6.5 x 10^-3 M NaOH Moles of NaOH: 0.0065 M = x / 0.015 L (0.0065) (0.015) = x x = 9.8 x 10^-5 mol NaOH Volume used: 15 mL + 35 mL = 50 mL Final concentration [NaOH]: 0.0065 M x 0.015 L / 0.050 L = Beaker E Initial concentration [NaOH]: 2.0 x 10^3 M NaOH Moles of NaOH: 0.0020 M = x / 0.010 L (0.0020) (0.010) = x x = 2.0 x 10^-5 mol NaOH Volume used: 10 mL + 40 mL = 50 mL Final concentration [NaOH]: 0.0020 M x 0.010 L / 0.050 L = Beaker F Initial concentration [NaOH]: 4.0 x 10^-4 M NaOH Moles of NaOH: 0.00040 M = x / 0.005 L (0.00040) (0.005) = x x = 2.0 x 10^-6 mol NaOH Volume used: 5 mL + 45 mL = 50 mL Final concentration [NaOH]: 0.00040 M x 0.005 L / 0.050 L = H 1. 2. Solution A: pH = 14.0 + log [0.013 M] = 12.1 ~ alkaline pH of NaOH Solution A according to pH paper: 11 pH difference: 12.1 - 11 = 1.1 Solution F: pH = 14.0 + log [0.00004 M] = 9.6 ~ less alkaline pH of NaOH Solution F according to pH paper: 7 pH difference: 9.6 - 7 = 4.4 I believe calculating the pH using itÕs concentration gives you more precise results because the chances of discrepancies are low in comparison to using pH paper. 3. Solution F has the highest concentration of H+ ions of 4.0 whereas Solution A has 1.3. [H+] = 10^9.6 [H+] = 4.0 [H+] = 10^12.1 [H+] = 1.3 1. Solution X: 10.00 mL - 1.0 M solution + 100 mL - total volume Solution Y: 5.0 mL of solution X + 95 mL of water Molarity = moles of solute / volume (L) Volume used: 10.00 mL + 100 mL = 110 mL ~ 0.11 L 1.0 M = x / 0.11 L (1.0) (0.11) = x x = 0.11 mol M = 0.11 mol / 0.11 L = 1.0 M 5.0 mL ~ 0.005 L C1 ¥ V1 = C2 ¥ V2 (1.0) (0.005) = (x) (0.0095) 0.005 = 0.0095x 0.005/0.0095 = 0.0095x/0.0095 0.053 M = x 2. 1) 0.050 L - 0.132 M NaOH + 2) 0.075 L - 0.120 M NaOH 3) Total molarity Molarity 1: 0.132 M = x / 0.050 L x = 0.066 mol NaOH Molarity 2: 0.120 M = x / 0.075 L x = 0.009 mol NaOH Molarity 3: Total amount of moles: 0.0066 + 0.009 = 0.0156 mol NaOH Total volume used: 0.050 L + 0.075 L = 0.125 L M = 0.0156 mol NaOH / 0.125 L = 1. What we know: ¥ Concentration of Solution A: 0.018 M ¥ 50.00 mL of NaOH solution ¥ Concentration of Solution F: 4.0 x 10^-5 M ¥ 1 drop = 0.05 mL What we need: ¥ The volume of Solution A needed to prepare the NaOH solution ¥ The amount of drops needed for Solution A [NaOH]a ¥ V.A = [NaOH]f ¥ V.F V.A = [NaOH]f x V.F / [NaOH]a V.A = 4.0 x 10^-5 M x 50.00 mL / 0.018 M = drops sol.A = 0.11 mL x 1 drop / 0.05 mL = 2.2 ~ 2 drops 2. I believe the more diluted the NaOH solution is, the lower the pH value. Because NaOH is a base, and in bases the pH will decrease with further dilution due to the decrease in OH- ion concentration. Conclusion To conclude, having performed the laboratory work, we learned the techniques used for serial dilution and how to calculate the concentration. The hypothesis was confirmed, having shown that each dilution decreases colour, concentration and pushes the final solution in beaker F to a pH of 7 a.k.a pure water. Moreover, the amount of moles do not change before and after dilution. And finally, we achieved the objective of creating several diluted sample solutions and discovering the pH of Solution A and F.