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Understanding Buffers and Titration Curves in Chemistry Lab

This lecture focuses on essential concepts in chemistry regarding buffers and titration curves. Students will complete lab reports, explore strong acid-strong base titrations, and calculate pH during titrations with various concentrations of acids and bases. Key topics include stoichiometric calculations, determination of pH at equivalence points, and application of the Henderson-Hasselbalch equation for buffer calculations. Join tutorials every Monday and Tuesday to enhance understanding and seek help during dedicated office hours.

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Understanding Buffers and Titration Curves in Chemistry Lab

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  1. Lecture #33 Chapter 17. Buffers CHEM100_11 This week: Labs. Reports should be completed in the lab. Bring calculators and rulers. Class average 63% Dr. Orlova PS-3026 Ph: 867-5237 gorlova@stfx.ca Student Chemical Society Tutorials: Every Monday and Tuesday , NH244 from 7pm-8:30pm Helping hrs at PS-3026: Tu & Th 1-2 pm; Wd 4-5 pm

  2. pH 14.0 11.0 10.0 9.0 8.0 pH=7 7.0 5.0 4.0 3.0 2.0 1.0 0.0 20mL mL NaOH • Strong Acid-Strong Base Titrations • A plot of pH versus volume of acid (or base) added is called a titration curve. • Lets add 20 mL of 0.1 M NaOH to 20mL of 0.1 M HCl. OH- Equivalence point NaCl + H2O H+ pHHCl = 1

  3. You are titrating 25.0 mL of 0.1 M HNO3 solution with 0.1M KOH. HNO3 (aq) + KOH (aq)  KNO3 (aq) + H2O (l) Calculate pH when: a) 24.9 mL of 0.1 M KOH have been added (HNO3 is still in excess) b) 25.1 mL of 0.1M KOH have been added (overtitrated- KOH is in excess)

  4. HNO3 (aq) + KOH (aq)  KNO3 (aq) + H2O (l) a) Moles H+ : (0.1mol/L x 0.0250 L) – (0.1mol/L x 0.0249) = 0.00001mol New volume: 0.0250L + 0.0249L = 0.0499L M H+ : 0.00001mol/ 0.0499 L = 0.0002 M pH = -log 2.0 x 10-4 =3.70

  5. b) Moles OH- : (0.1mol/L x 0.0251 L) – (0.1mol/L x 0.0250) = 0.00001mol M OH- : 0.00001mol/ 0.0501 L = 0.0001996 M pOH = -log 1.996 x 10-4 =3.70 pH = 14-3.70 = 10.3 Useful information: 1mL = ca. 20 drops; 1 drop = 0.05mL In the vicinity of the equivalence point, 4 drops of 0.1 M KOH changed pH from 3.7 to 10.3

  6. pH 14.0 11.0 10.0 9.0 8.0 7.0 5.0 4.0 3.0 2.0 1.0 0.0 20mL mL NaOH Titration of a weak acid with a strong base 20 mL of 0.1M solution of acetic acid is titrated with 0.1 M solution of NaOH Ka = 1.8 x 10-5

  7. pH 14.0 11.0 10.0 9.0 8.0 7.0 5.0 4.0 3.0 2.0 1.0 0.0 20mL mL NaOH OH- Equivalence point pH > H+

  8. pH before titration: Ka = [H+][CH3COO-]/[CH3COOH] X2/0.1 = 1.8 x 10-5 ; X2 = 1.8 x 10-6 X= √1.8 x 10-3 = 1.3 x 10-3 pH = -log 1.3 x 10-3 = 2.89

  9. pH between initial and equivalent points CH3COOH (aq) + NaOH (aq)  CH3COONa(aq) + H2O (l) CH3COOH (aq) + OH- (aq)  CH3COO- (aq) + H2O (l) BUFFER! To find pH of the buffer, make 2 steps: 1. Consider neutralization rxn and using stoichiometric calculations find concentrations of CH3COOH and CH3COO- in the solution 2. Consider buffer’s equilibrium and find [H+] using reaction table or Henderson- Hasselbalch equation

  10. Calculate pH when 18 mL of NaOH have been added to 20 mL of 0.1M CH3COOH (Ka = 1.8 x 10-5; pKa = 4.74) 1. Stoichiometric calculations Moles AH : 0.1 mol/L x 0.020L = 0.0020 mol Moles OH- : 0.1 mol/L x 0.018L = 0.0018 mol Before rxn 0.0020 0.0018 0 CH3COOH (aq) + OH- (aq)  CH3COO- (aq) + H2O (l) After rxn 0.0002 0 0.0018

  11. But because [H+] = Ka [HA]/[A-] Ratio of molarities is the same as ratio of moles We can use moles Formally, we should use molarities Molarities: 0.0002/0.038 = 0.0053: 0.0018/0.038 =0.047 Same is valid for Henderson-Hasselbalch pH = pKa + log [CH3COO-]/[CH3COOH] pH = 4.74 + log (0.0018/0.0002) =4.74 + 0.95 = 5.69

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