Buffering Regions Endpoints and Indicators The endpoint is

Buffering Regions, Endpoints and Indicators The endpoint is the observable colour change. The equivalence point is when the amount of acid and base are precisely chemically equal. NO!!

Before any acid titrant is added, the sample solution is mostly water molecules and excess hydroxide ions. Buffering occurs because initially, any acid added immediately reacts with the excess hydroxide and is converted to water. NO!!

The equivalence point is approached as the excess hydroxide ions in the sample solution are almost all reacted with the added acid. Now the solution consists of water molecules and excess hydronium ions. NO!!

Bromothymol blue is a good indicator to use for a strong acid–strong base titration, because it changes colour very close to the equivalence point.

Acid–Base Indicator Equilibrium An indicator is a Brønstead–Lowry conjugate weak acid–base pair formed when an indicator dye dissolves in water. conjugate pair HIn(aq) + H 2 O(l) ↔ In-(aq) + H 3 O+(aq) Acid (red litmus) Base (blue litmus) An increase in hydronium ions present causes an equilibrium shift to the left, producing more of the red–coloured acid form. The presence of excess hydroxide, the hydroxide removes hydronium, producing an equilibrium shift to the right (blue colour).

Sample Problem • Use the following p. H scale to label the colors for bromothymol blue over the 0 -14 p. H range. Identify the form of the indicator for each distinct color using conventional symbols. 0 7 14

Sample Problem: Complete the analysis section of this investigation. • Problem: What is the approximate p. H (i. e. p. H range) of an unknown solution? • Evidence: Separate samples of the unknown solution turned blue litmus to red, congo red to blue and orange IV to yellow. • Analysis:

Sample Problem • Three unknown solutions in unlabelled beakers have p. H values of 5. 8, 7. 8, and 9. 8. Write two diagnostic tests using indicators to identify the p. H of each solution.

Sample Diploma Question A sample of rain water is poured into five test tubes. A different indicator is added to each test tube. Four of the observations are recorded in the table below. Color Indicator Methyl red Yellow Phenol red Yellow Bromocresol green Phenolphthalein Bromothymol blue Blue Colorless ? The p. H of the rainwater and the predicted color of the sample containing bromothymol blue are: A. 6. 0 and blue B. 7. 6 and blue C. 6. 0 and yellow D. 7. 6 and yellow

Sample Diploma Question Pickling vinegar has a p. H of 2. 37. When 3 drops of bromocresol green and 3 drops of phenolphthalein are added to a sample of this vinegar, the resulting color of the solution is: A. Yellow B. Green C. Blue D. Purple

Sample Diploma Question The labels came off four cleaning solution containers found under a kitchen sink. Each of the cleaning solutions was tested with two available indicators and the following results were recorded. Cleaning Solution Bromothymol blue Phenolphthalein 1 Blue Pink 2 Blue Colorless 3 Green Colorless 4 Blue Light Pink Listed in order from lowest to highest p. H, the cleaning solution are, respectively, , and. (Record your four digit answer in the numerical response section on the answer sheet)

Sample Diploma Question A. B. C. D. The amphiprotic species that reacts with bromothymol blue to produce a yellow color is: Na. HSO 4(aq) Na. HCO 3(aq) Na. OCl(aq) H 2 O(l)

Polyprotic Entities and Sequential Reactions Na 2 CO 3(aq) → 2 Na+(aq) + CO 32–(aq) The carbonate ion is a diprotic base because it can accept two protons. Hydrochloric acid is a strong acid, so it is like adding hydronium ions (H 3 O+(aq)).

H 3 O+(aq) + CO 32–(aq) → H 2 O(l) + HCO 3–(aq) H 3 O+(aq) + HCO 3–(aq) → H 2 O(l) + H 2 CO 3(aq)

Nitric acid is a monoprotic acid: HNO 3(aq) — NO 3–(aq) 1 hydrogen! Sulfuric acid is a diprotic acid: H 2 SO 4(aq) — HSO 4–(aq) — SO 42–(aq) 2 hydrogens! Remember: for every proton transferred by a polyprotic entity, the strength of the new acid or base entity formed greatly decreases.

H 3 PO 4(aq) — H 2 PO 4–(aq) — HPO 42–(aq) — PO 43–(aq) OH–(aq) + H 2 PO 4– (aq) → H 2 O(l) + HPO 42– (aq) OH–(aq) + H 3 PO 4(aq) → H 2 O(l) + H 2 PO 4– (aq) As a general rule, only quantitative reactions (i. e. 100% reaction) produce detectable equivalence points in an acid–base titration.

Sulfuric Acid- A Unique Polyprotic Acid • Sulfuric acid (H 2 SO 4(aq)) is unique because it is the only strong acid that is polyprotic, meaning that the first proton lost is already quantitative in an aqueous solution. • As a result, its 100% ionization produces hydrogen sulfate ions (HSO 4(aq)) as well as hydronium ions. – The hydrogen sulfate ion is the ONLY hydrogen polyatomic that is a weaker base than water, so it WILL NOT act as a base in aqueous solutions. – As an acid, however, it will USUALLY react quantitatively with a base (assuming the base is in excess) in two complete proton transfers.

Sample Problem Use the accompanying sketch of a p. H curve for a titration to answer the following questions. 1. Does the burette contain the acid or the base? 2. Is the sample reacted an acid or a base? 3. How many endpoints are represent? 4. Estimate each endpoint. 5. How many quantitative reactions have occurred? 6. Choose the best indicator for each endpoint. 7. What part of the curve represents a possible buffering region? Acid-Base Reaction Volume (m. L)

ü Read pgs. 751 – 759 ü pgs. 754, 755, 759 Practice #’s 1 – 10

p. H Curve Shape versus Acid and Base Strength Weak acid–weak base titrations do not have a detectable equivalence point, so p. H curves are normally not done for those reactions.

Strong Base/Strong Acid H 3 O+(aq) + OH–(aq) → 2 H 2 O(l) • equivalence point is always at a p. H of 7 • large vertical portion • a wide range of indicators would be useful

Strong Acid/Weak Base For example, hydrochloric acid is added to aqueous ammonia: H 3 O+(aq) + NH 3(aq) → H 2 O(l) + NH 4+(aq) • equivalence point is always less than a p. H of 7 equivalence point p. H = 4. 6 • choose an indicator that changes colour at a lower p. H value e. g. bromocresol green

Strong Base/Weak Acid For example, barium hydroxide is added to acetic acid: CH 3 COOH(aq) + OH–(aq) → H 2 O(l) + CH 3 COO –(aq) equivalence point p. H = 9. 2 • equivalence point is always greater than a p. H of 7 • choose an indicator that changes colour at a higher p. H value e. g. phenolphthalein

How Buffers Work • Buffers resist changes in p. H if the quantity of excess acid or base added is less than the quantities of the conjugate pair entities present in the buffer. – When an acid is added, the basic entity of the conjugate acid-base pair neutralizes the acid until consumption is complete and the acid is in excess resulting in a dramatic p. H fall. – Likewise, when a base is added, the acidic entity of the conjugate acid-base pair neutralizes the base until consumption is complete and the base is in excess resulting in a dramatic p. H increase. • Buffering Capacity refers to the limit of the ability of a buffer to maintain a p. H level. When the entity of the conjugate acid-base pair that reacts with the added reagent is completely consumed. The buffering fails and the p. H changes drastically. (See Figure 16 on p. 764)

p. H Curve Buffering Regions and Buffering Solutions A buffer is a combination of any weak acid with its conjugate base, in the same solution.

(a) OH–(aq) + CH 3 COOH(aq) → H 2 O(l) + CH 3 COO–(aq) (b) H 3 O+(aq) + CH 3 COO–(aq) → H 2 O(l) + CH 3 COOH(aq) Buffer capacity is the limit of the ability of a buffer to maintain a p. H level.

Blood plasma is buffered because the chemical reactions that take place in our bodies must be in a narrow p. H range.

Sample Diploma Question The main buffer solution of plasma and tissue fluid found in our bodies is H 2 CO 3(aq) - HCO 3 -(aq). When excess hydronium ions enter our blood, the equation that represents the reaction that occurs is: A. B. C. D. H 3 O+(aq) + OH-(aq) 2 H 2 O(l) H 2 CO 3(aq) + OH-(aq) HCO 3 -(aq) + H 2 O(l) H 2 CO 3(aq) + H 2 O(l) H 3 O+(aq) + HCO 3 -(aq) H 2 CO 3(aq) + H 2 O(l)

Sample Diploma Question Blood maintain a nearly constant p. H because it contains: A. B. C. D. sodium ions and chloride ions that keep the p. H of the blood at 7 hemoglobin that maintains the oxygen levels in the blood catalysts (enzymes) that control the equilibrium in the blood buffers that regulate the hydronium ion concentration in the blood

Sample Diploma Question Which of the following equimolar solutions could act as a buffer system? (40% of students got this WRONG on the diploma!) A. B. C. D. KH 2 PO 4(aq) /H 3 PO 4(aq) KCl(aq)/HCl(aq) KCl. O 4(aq)/HCl. O 4(aq) KNO 3(aq)/HNO 3(aq)

ü Read pgs. 760 – 766 ü pgs. 762, 763, 766 Practice #’s 11 – 21 ü pg. 767 Section 16. 4 Questions 1 – 10

Acid- Base Stoichiometry • When completing stoichiometric calculations from titration evidence, the number of quantitative reactions that occur is determined by the p. H curve. The number of reactions is equal to the number of protons transferred which, is equal to the number of endpoints. • In acid-base stoichiometry two different methods that can be used to determine the net equation. 1. Using a non-ionic equation 2. Using the method of predicting acid-base reactions

Method 1: Using a non-ionic equation (double replacement reaction) to determine a balanced chemical equation • Method 1 is often simpler and more convenient to use, however it requires careful consideration of the mole ratio since polyprotic substances can transfer more than one proton. ** This method is best if the base can accept ALL protons. For example, carbonic acid (H 2 CO 3(aq)) reacts with sodium hydroxide to the second endpoint.

Method 2: Using the method of predicting acid-base reactions to determine the net ionic equation • Method 2 is best used when not all the protons are being transferred from the acid to the base. For example, phosphoric acid (H 3 PO 4(aq)) reacts with sodium hydroxide until a second endpoint is observed. • However, regardless of the method used, the resulting equation should be the same! • The remaining process of solving acid-base stoichiometry remains the same as with any stoichiometry problem.

Example An investigation is preformed to determine the concentration of oxalic acid in a rust removing solution. Three 10 m. L samples of oxalic acid are titrated with a standardized 1. 27 mol/L Na. OH(aq) solution. A phenolphthalein indicator is used as determine the second endpoint. The student collected the following evidence. Trials 1 2 3 Final buret reading (m. L) 11. 6 23. 1 34. 5 Initial buret reading (m. L) 0. 0 11. 6 23. 1 Volume of Na. OH(aq) added (m. L) 11. 6 11. 5 11. 4 Determine the concentration of oxalic acid.

Solution using Method 1 ** 2 endpoints indicate that two protons were transferred. This means that a compound containing the oxalate ion must be a product. HOOCCOOH(aq) + 2 Na. OH(aq) Na 2 OOCCOO(aq) + 2 H 2 O(l) To solve: 1. Balance chemical equation including states. List given quantities underneath the appropriate formula. HOOCCOOH(aq) + 2 Na. OH (aq) 2 HOH(l) + Na 2 OOCCOO(aq) V= 0. 0100 L v= 0. 0115 L C =? C= 1. 27 mol/L 2. Find number of moles n = cv n= (0. 0115 L)(1. 27 mol/L) = 0. 0146… mol of OH-(aq) 3. Mole-mole ratio 0. 0146 mol of OH-(aq) X 1 mol HOOCCOOH(aq) = 0. 00730 mol 2 mol OH-(aq) 4. Solve C = n = 0. 00730 mol = 0. 730 mol/L V 0. 010 L

Solution using Method 2 1 st endpoint: HOOCCOOH(aq), Na+(aq), OH-(aq), H 2 O(l) SA SS SB A/B Chemical equation: HOOCCOOH(aq) + OH-(aq) HOH(l) + HOOCCOO-(aq) 2 nd endpoint: continue to add more strong base Chemical entities: HOOCCOO-(aq), Na+(aq), OH-(aq), H 2 O(l) SA SS SB A/B Chemical equation: HOOCCOO-(aq) + OH-(aq) HOH(l) + OOCCOO 2 - (aq) Net equation: HOOCCOOH(aq) + OH-(aq) HOH(l) + HOOCCOO-(aq) + OH-(aq) HOH(l) + OOCCOO 2 - (aq) HOOCCOOH(aq) + 2 OH- (aq) 2 HOH(l) + OOCCOO 2 - (aq) NOTE: When the spectator species from method one (Na+(aq)) is eliminated, both methods result in the same equation!

Sample Problem • A sodium hydrogen phosphate solution is titrated with hydrochloric acid. Only one quantitative reaction is observed. Sketch the p. H curve and provide the relevant equilibrium equation.

Sample Diploma Question • Determine the concentration of phosphoric acid when 10. 0 m. L is titrated with a 0. 250 mol/L potassium hydroxide solution to the second endpoint. The volume of KOH(aq) that was used to reach this endpoint is 15. 0 m. L. (answer: 0. 188 mol/L)

Sample Diploma Question Use the following information to answer the next two questions A student titrated a 10. 0 m. L sample of nitric acid with a 0. 00500 mol/L sodium hydroxide solution in the presence of an indicator. The results were as follows: Trials 1 2 3 4 Final buret reading (m. L) 7. 99 14. 51 21. 02 27. 53 Initial buret reading (m. L) 1. 00 7. 99 14. 51 21. 02 Volume added (m. L) 6. 99 6. 52 6. 51 1. The average volume of titrant used is m. L (Record your answer using three digits) (answer: 6. 51 m. L) Use the recorded value from numerical response 1 to answer numerical response 2. 2. The concentration of the nitric acid is mmol/L (Record your answer using three digits) (answer: 3. 26 mmol/L)

Acid- Base Stoichiometry Worksheet 1. 2. In a chemical analysis, 25. 0 m. L of sulfuric acid is titrated to the second endpoint with 0. 358 mol/L KOH(aq). In the titration, an average volume of 18. 2 m. L was required. Calculate the molar concentration of the sulfuric acid. (0. 130 mol/L) Several 10. 0 m. L vinegar samples were titrated with a standardized 0. 582 mol/L solution of sodium hydroxide. An average volume of 13. 8 m. L of sodium hydroxide was required to reach the phenolphthalein endpoint. What is the concentration of the vinegar solution? (0. 803 mol/L)

3. A sodium borate solution was titrated to the second endpoint with 0. 241 mol/L hydrobromic acid. An average volume of 15. 2 m. L of hydrobromic acid was required to react with 20. 0 m. L samples of sodium borate. Calculate the molar concentration of sodium borate. (0. 0916 mol/L) 4. Complete the analysis for the following investigation report Problem: What is the concentration of a hydrochloric acid solution? (answer: 0. 179 mol/L) Experimental Design: 100. 0 m. L of a standard solution of sodium oxalate was prepared using 1. 85 g of the dry solid. Using the second endpoint, 10. 0 m. L samples were titrated with hydrochloric acid. Evidence: Trials 1 2 3 4 Final buret reading (m. L) 16. 1 31. 5 46. 9 16. 9 Initial buret reading (m. L) 0. 3 16. 1 31. 5 Volume added (m. L)