Lab Report Exp. 15 PDF

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Lab Report Experiment 15. pH and pH Titrations...

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4/5/18 Experiment 15. pH and pH Titrations

Purpose: The purpose of this experiment is to notice the different rates of pH change at different stages of the titration of solutions of weak acids. This is needed because we would need to determine the pKa of our benzoic acid which is by measuring the pH of the solution at the half- equivalence point. Theory/Principles: Goldwhite, H.; Tikkanen, W. Experiment 15. pH and pH Titrations, Experiments in General Chemistry, 4th ed.;The McGraw Hill Companies. (97-98) Experimental Procedures: Goldwhite, H.; Tikkanen, W. Experiment 15. pH and pH Titrations, Experiments in General Chemistry, 4th ed.;The McGraw Hill Companies. (98-99) Data Tables/ Summary: -

Table #1: Raw Data collected for parts A/B & signed pre-lab

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Table #2: Formatted Data for part A of experiment & calculated expected pH

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Table #3: Formatted Data for mass of beaker & mass of beaker + benzoic acid for part B of experiment

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Table #4: Formatted Data for Titration B#1 with increments of 1.0 mL along with its graph of pH vs. NaOH volume

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Table #5: Formatted Data for titration B# 2 with increments of 0.2 mL, 1.0 mL, and 0.1 mL along with its graph of pH vs. NaOH volume

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Table #6: Equilibrium point, half-equivalence point, accepted benzoic acid Ka, actual benzoic acid MM, and calculated equivalent weight of benzoic acid along with its % error

Results & Discussions: Sample Calculations: Part A: 0.015 M HCl, fully dissociates (strong acid) pH=-log(0.015 M)= 1.82 0.06 M HOAc, 0.04 M NaOAc Ka= 4 × 10^-5 pH= -log(4×10^-5) + log(0.04/0.06) =4.22 0.022 M NaOH, fully dissociates (strong base)

pOH=-log(0.022)=1.66 pH=14.00-1.66=12.34 0.1 M NH4Cl R

NH4 (aq) + H2O (l) ⇌ H3O+ (aq) + NH3 (aq)

I(M) 0.1

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0

0

C(M) -x

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x

x

E(M) 0.1-x

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x

x

Kb NH3= 1.8 ×10^-5 Ka= kw/kb = 1.0 ×10^-14/1.8×10^-5 = 5.6 × 10^-10 Ka= [H3O][NH3]/[NH4] = x^2/0.1-x = 5.6 × 10^-10 √x^2= √5.6×10^-11 = 7.48 ×10^-6 -log(7.48×10^-6)= 5.13 0.1 M NaOAc Just like above set up RICE table Ka acetic acid= 1.86 × 10^-5 Kb= 1.0×10^-14/1.86 × 10^-5 = 5.38 ×10^-10 Kb= x^2/0.1-x = 5.38 ×10^-10 √x^2=√5.38×10^-11 = 7.34 ×10^-6 -log(7.34 ×10^-6)= 5.13 pH=14.00-5.13= 8.87 0.015 M NH4Cl with 0.02 M NH3 Kb= 1.8 ×10^-5 -log(1.8 ×10^-5)= 4.74

pOH= 4.74 + log(0.015/0.02) = 4.62 pH=14.00-4.62= 9.38 Part B: Equilibrium point: between 22.84 mL-22.95 mL pH=pKa Half-Equivalence point: 11.52 mL pH= pKa= 4.72 ka= 10^-4.72 = 1.91 × 10^-5 Accepted benzoic acid Ka= 6.5 × 10^-5 Actual benzoic acid Molar Mass: 122.12 g/mol MNaOH= 0.10 M VNaOH= 22.84 mL (closer to neutralization) V Benzoic Acid= 50 mL M Benzoic Acid= 0.10 × 22.84/50 = 0.046 M Moles of Benzoic Acid= 0.046 M × 0.050 L= 0.0023 moles MW= 0.30 g/0.0023 moles= 130.43 g/mol (for B#2) % error of MW Benzoic acid= 130.43-122.12/122.12 ×100 = 6.8% Logical Explanation: For part A of the experiment we were to measure and record the pH of the solutions given with a pH meter. With our collected pH measurements and calculated expected pH of each of these solutions we can compare our measured pH values with the calculated. Based off our data and calculations we notice that our measured pH values are pretty consistent with the expected pH values. However, for 0.06 M HOAc and 0.04 M NaOAc we made the error of not mixing the

both together and measuring its pH value, instead we got the pH value for each. For part B of the experiment for titration of our benzoic acid #1 with increments of 1.0 mL we noticed the pH slowly rising with each mL. As our titration continues I believe there must have been a malfunction with our pH meter has our pH values aren’t increasing or showing much different within titrating. Eventually, we reached its peak at 27.8 mL with a pH value of 11.18. For our next titration for our benzoic acid #2 with increments of 0.2 mL, 1.0 mL, and 0.1 mL we see pH values start to slowly increase as well. For this titration we are able to determine the equivalence point and half equivalence point by looking at our plot of pH vs. NaOH volume above (Table #5). The equivalence point is where neutralization occurs which from our data it must have occurred between 22.84 mL-22.95 mL. We couldn’t accurately pinpoint where neutralization occurred because we must have overshot the titration because at these two points we have pH values of 6.54 and 8.42. However, given that 22.84 mL has the pH value closest to the neutralization of 7, I went ahead in using that measurement to help me determine the equivalent molar mass of our benzoic acid. Going ahead with calculating our equivalent weigh of benzoic acid we ended up getting 130.43 g/mol while the actual mass of benzoic acid is 122.12 g/mol. As we notice our equivalent mass of benzoic acid is not much of a difference of its actual mass along with a % error of only 6.80 %. Now for comparing our experimental Ka value of our benzoic acid which is 1.91 x 10^-5 with the accepted Ka being 6.5 x 10^-5 we notice a huge difference between the two. The big difference between could have been due to errors in titration such as overshooting, mixing up increments, or pH meter malfunction. Discussion Ques.:

1. We cannot prepare a solution of pH= 8 by diluting sufficiently 0.01 M solution of HCl with water because with higher acid or base concentrations then water will be neglected unless acid or base concentrations are lower. 2. Our graphical method used in our experiment for determining Ka is a better way because we can see the pH changes occurring with each mL increment during titration. With this method we can then accurately predict Ka for benzoic acid. 3. I believe using methyl orange as an indicator would be better because we would find bigger value of molecular weight. Moles NaOH: 0.10 M x 0.61 mL = 0.061 moles MW: 0.30g / 0.061 moles= 4.91 g/mols (methyl orange/ I used my second titration data for calculation) Moles NaOH: 0.10 M x 23.01 mL= 2.301 moles MW: 0.30g / 2.301 moles= 0.13 g/mols (phenolptalein) 4. % unc. in g BA= 0.0001 g / 0.30 g x 100 % = 0.03 % % unc. Veq. = ±0.10 mL/ 22.84 x 100 %= 0.44 % % unc. [NaOH]= 0.01 % % unc. in MW BA= 0.03 % + 0.44 % + 0.01 %= 0.48 %

Conclusion:

By carrying out titrations of solutions of weak acids and taking notice of the different pH changes occurring at different rates we were able to determine the pKa of our benzoic acid at the half-equivalence point to be 1.91 x 10^-5. References: Goldwhite, H.; Tikkanen, W. Experiment 15. pH and pH Titrations, Experiments in General Chemistry, 4th ed.;The McGraw Hill Companies. (98-100) Gilbert, T. R.; Kirss, R. V.; Foster, N.; Bretz, S. L.; Davies, G. Chemistry: The Science in Context, 5th ed.; W.W. Norton & Company, Inc: New York, 2017....