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Analytical Chemistry (Volummetric analysis) Lecture 4 Prepared by Dr.Ahmed Bahgat Example 9: What is the pH of a solution made by dissolving 2.5 g NaOH in 400 mL water? Determine number of moles of NaOH x mol of NaOH = 0.0625 mol NaOH...

Analytical Chemistry (Volummetric analysis) Lecture 4 Prepared by Dr.Ahmed Bahgat Example 9: What is the pH of a solution made by dissolving 2.5 g NaOH in 400 mL water? Determine number of moles of NaOH x mol of NaOH = 0.0625 mol NaOH Calculate the molarity of the solution [Recall 1000 mL = 1 L] MNaOH = 0.15625 molar NaOH Na+ + OH- 0.15625 molar 0.15625 molar 0.15625 molar pOH = -log [OH-] or kW = [H+] [OH-] pOH = -log [0.15625 M] 1 x 10-14 = [H+] [0.15625 M] pOH = 0.8 [H+] = 6.4 x 10-14 M pOH + pH = 14 pH = -log [H+] 0.8 + pH = 14 pH = 13.2 pH = -log [6.4 x 10-14 M]  Neutralisation Titrations, Acid – Base Titrations Acid–Base titration: A titration in which the reaction between the analyte and titrant is an acid–base reaction. Calculation of titration: M: Molarity + - moles H3O = moles OH V: volume n: # of H+ ions in the acid MVn = MVn or OH- ions in the base Example: 10 ml of NaOH was titrated against H2SO4 (0.1M), the end point appeared after adding 9 ml of H2SO4. 1- write the balanced chemical equation. 2- Calculate the concentration of NaOH Answer: 1- The balanced chemical equation: 2 NaOH + H2SO4 → Na2SO4 + 2H2O 2- The Molarity of NaOH: moles H2SO4 = moles NaOH M×V× n = M×V× n 0.1×9×2 = ? ×10×1 Molarity of NaOH= 0.18 mol/L  Titrations and pH curves (cont.) In this section, we will discuss the changes that occur in pH during an acid base titration. The progress of an acid-base titration is often monitored by plotting the pH of the solution being analyzed as a function of the amount of titrant added. Such a plot is called a pH curve or a titration curve. Titration curve: is a plot of pH vs. the volume or the amount of titrant added.  Titration of Strong Acid against Strong Base Strong acid- strong base are completely dissociated in aqueous solution. The net ionic reaction for a strong acid-strong base titration is H+(aq) + OH-(aq) H2O(l) - The pH at various points during a titration can be calculated directly from the stoichiometric amounts of acid and base that have been allowed to react. To compute [H+] at a given point in the titration, we must determine the amount of H+ that remains at that point and divide by the total volume of solution. Since titrations usually involve small quantities (burettes are graduated in milliliters) we are going to use the mmoles = volume (in ml) X molarity.  Titration of a Strong Acid With a Strong Base Titration of a Strong Acid With a Strong Base Solution 14.0 of NaOH OH - Na+ Na+ 12.0 OH- OH- Na+ Na+ 10.0 OH- 8.0 pH equivalence point Solution 6.0 of HCl H+ Cl Cl- H+ H+ 4.0 H+ Cl- Cl- 2.0 0.0 0.0 10.0 20.0 30.0 40.0 Volume of 0.100 M NaOH added (mL) 1- Adding NaOH from the burete to hydrochloric acid in the flask. In the beginning the pH increases very slowly. 2- Adding additional NaOH is added. pH rises as the equivalence point is approached. 3- Change occurs near the equivalence point, then the addition of a small amount of OH- produces a large change in the pH value..  Example: - Consider the titration of 50.0 ml of 0.1 M HCl with 0.1 M NaOH. Calculate the pH of the solution after the addition of 10.0, 40, 50.0 and 60.0 ml of 0.1M NaOH. a. Initial pH. HCl is a strong acid and is completely dissociated [H+] = 0.1 M and pH = -log 0.1 =1 b. 10.0 ml of 0.1M NaOH has been added. H+ + OH- H2O Before reaction (50.0 x 0.1) (10 x 0.1) = 5.0 mmol 1.0 mmol 5.0 - 1.0 1.0 - 1.0 After reaction = 4.0 mmol = 0 After reaction, + [H ] = no. of mmol HCl remained/total no. of mmol [H+] = 4.00m mol/60.0 ml= 6.67 x 10-2 mmol/ml pH= 2- log 6.67 =1.18  (cont.) c. pH at the equivilence point. 50 ml (total) of 0.1 M NaOH has been added. H+ + OH- H2O - Before 50.0 x 0.1 = (50 x 0.1) - Reaction = 5.0 mmol = 5.0 mmol - Change = -5.00 -5.00 The equilibrium is: 2H2O ↔ H3O+ + OH- and [H3O+] [OH-] = Kw = 1.0 x 10-14 Since [H3O+] = [OH-] = [H3O+]2 = 1 x 10-14 [H3O+] = 1 x 10-7 pH = 7  (cont.) d. 60 ml of 0.1m NaOH has been added H+ + OH- H 2O Initial 50.0x0.1 60x0.1 = 5.0 mmol = 6.0 mmol Change -5.0 -5.0 At equilibrium - -1.0 The OH- ion concentration is [OH-] = 1.00 mmol/ 110 ml = 9.1 x 10-3 M pOH = 3- log 9.1 = 2.04 pH = 14 – 2.04 = 11.96  (cont.) The results of the previous steps are summarized by the pH curve shown in fig. Note that the pH changes very gradually until the titration is close to the equivalence point, where a dramatic change occurs. This behavior is due to the fact that early in the titration, there is relatively a large amount of H+ in the solution, and the addition of a given amount of OH- produces a small change in pH. However, near the equivalence point [H+] is relatively small and the addition of a small amount of OH- produces a large change.  Titrations of weak acids with strong bases For strong acids and strong bases titrations, as they are completely dissociated, the calculations to obtain the pH curves for titrations are quite straightforward. When the acid being titrated is a weak acid, there is a major difference: to calculate [H+] after a certain amount of strong base has been added, we must deal with the weak acid dissociation equilibrium. Titration of a Weak Acid With a Strong Base Titration of a Weak Acid With a Strong Base Titration Data 14.0 NaOH added 12.0 (mL) pH 10.0 0.00 2.89 equivalence point 5.00 4.14 8.0 pH 10.00 4.57 12.50 4.74 6.0 15.00 4.92 4.0 20.00 5.35 24.00 6.12 2.0 25.00 8.72 26.00 11.30 0.0 30.00 11.96 0.0 10.0 20.0 30.0 40.0 40.00 12.36 Volume of 0.100 M NaOH added (mL) Phenolphthalein is best indicator: pH change 8.0 - 9.6  Example Consider the titration of 50.0 ml of 0.1M acetic acid (HC2H3O2, Ka = 1.8x10-5) with 0.1 M NaOH. As before we will calculate the pH at various points representing volumes of added NaOH. Before doing anything we have to calculate the volume of NaOH required for complete neutralization, (MV)acetic = (MV)NaOH (50x0.1)acetic = (0.1V)NaOH  VNaOH= 50 ml - A. No NaOH has been added. - Since HB is a weakly dissociated base This is a typical weak acid calculation and the pH = 2.87 HB + H2O ↔ H3O+ + B- (0.1-x) x x We assume that [H3O+ ] = [B-] Ka = 1.0 × 10-5 Ka = [H3O+ ] [B-] / [HB] = [H3O+ ]2 / 0.1 = 1.0 × 10-5 [H3O+ ] = 1.0 × 10-3 pH = 3.00  Sample Calculation: Weak Acid-Strong Base Titration Curve (Cont.I) Region 2. After 10. mL of base (total) has been added. This the buffer region of the titration curve.. So pH = pKa + log [conj. Base] [acid]. [H+]= Ka [HB] / [B-] -log[H+]= -log Ka - log[HB] /[B-] ‫؞‬ pH = pKa + log [B-]/[HB] But first must calculate the nominal amounts of acid and base forms of the weak acid created by addition of the strong base. These are: [HB]0 = [B-]0 Example (cont.) B. 10.0 ml of 0.10 M NaOH has been added The major species in the mixed solution before any reaction occur are HC2H3O2, OH- , Na+ and H2O. The stoichiometry problem c. pH at equivalence point. mmol HB + OH- → B- + H2O Initial 5.0 5.0 - Change -5.00 -5.00 +5.00 Equilibrium - - 5.00 B- + H2O ↔ HB + OH- [HB] [OH-] = Kb = Kw = 1.0 × 10-14 = 1 × 10-9 [B-] Ka 1.0 × 10-5 [B-] = 5.00/100 * Since the dissociation produces 1 HB and 1 OH-, so [HB] ≈ [OH-] Then [OH-]2/ 0.05 = 1.0 × 10-9 [OH-] = 7.1 × 10-6 pOH = 5.15 and pH = 8.85  (example cont.) E. 60.0 ml (total) of 0.1 M NaOH has been added. At this point excess OH- has been added. The stoichiometry will be After the reaction is complete, the solution contains the major species Na+, C2H3O2-, OH- and H2O There are two bases in this solution C2H3O2-, OH-. However, C2H3O2-,is a weak base compared with OH-. Therefore the amount of OH- produced by C2H3O2- with H2O will be small compared with the excess OH-. The pH can be calculated as:  (example cont.) This is an important result: the pH at the equivalence point of a weak acid with a strong base is always greater than 7. This is because the anion of the acid that remains at the equivalence point is a base, In contrast to the case of strong acid- strong base titration, the pH is 7 because the anion remaining in this case is not an effective base. Comparing the titration curve with that of strong acid-strong base titration, notice that: Near the beginning of the titration of the weak acid, the pH increases more rapidly than it does in the strong acid case. The other notable difference between the curves for strong and weak acids, is the value of the pH at the equivalence point. *For strong acids, pH = 7 and for weak acids, pH is greater than 7 (alkaline). The four Major Differences Between a Strong Acid- Strong Base Titration Curve and a Weak Acid-Strong Base Titration Curve 1. The initial pH is higher. 2. A gradually rising portion of the curve, called the buffer region, appears before the steep rise to the equivalence point. 3. The pH at the equivalence point is greater than 7.00. 4. The steep rise interval is less pronounced.  Notice that: The equivalence point in an acid-base titration is defined by the stoichiometry and not by the pH. The equivalence point occurs when enough titrant has been added to react exactly with the acid or base being titrated.  Example Hydrocyanic acid (HCN), is very weak acid (Ka= 6.2x10-10) when dissolved in water. If a 50.0 ml sample of 0.10 M HCN is titrated with 0.10 M NaOH, calculate the pH of the solution at the equivalence point of the titration. At the equivalence point The equivalence point will occur when a total of 5 mmol OH- has been added. Since the NaOH solution is 0.1M, the equivalence point will occur when 50 ml NaOH are added.  Example (cont.) The major species are CN-, Na+ and H2O, thus the reaction that will control the pH involves the basic cyanide ion extracting a proton from water: CN-(aq) + H2O(l) → HCN(aq) + OH-(aq)  Thus [CN-] = 5.00/100 = 0.05 M CN-(aq) + H2O(l) → HCN(aq) + OH-(aq) * Since the dissociation produces 1 mole HCN and 1 mole OH-, so [HCN] ≈ [OH-] Then [OH-]2/ 0.05 = 1.6 × 10-5 [OH-] = 8.9x10-4 Then, from Kw we have [H+] = 1.1x10-11 and pH = 10.96  Titrations of weak bases with strong acids Titrations of weak bases with strong acids Can be treated using the same procedures as before. Always, you should think first about the major species in solution and decide whether a reaction occurs that runs to completion. Run stoichiometry and equilibrium calculations a to get the pH. Titration of a weak base & a strong acids Because the reactants are a weak base and a strong acid, the reaction goes essentially to completion after each addition of acid. 1. Before any acid is added, the solution contains only the weak base B + H2O ↔ BH+ + OH- Kb 2. Between the initial point and the equivalence point: buffer! 3. At the equivalence point 4. Beyond the equivalence point: the pH is determined by the excess acid Titration of a Weak Base With a Strong Acid Titration of a Weak Base With a Strong Acid Titration Data 14.0 HCl added 12.0 (mL) pH 10.0 0.00 11.24 10.00 9.91 8.0 pH 20.00 9.47 6.0 30.00 8.93 equivalence point 40.00 8.61 4.0 45.00 8.30 47.00 7.92 2.0 48.00 7.70 49.00 7.47 0.0 0.0 10.0 20.0 30.0 40.0 50.0 50.00 5.85 51.00 3.34 Volume of 0.100 M HCl added (mL) - Titrations of 50 ml weak bases (NH3 0.1M) with 0.1M HCl (cont.) Before the addition of any HCl Major species: NH3 and H2O This is a case of weak base. NH3 + H 2O ↔ NH4+ + OH- (0.1- X) X X Kb = [(NH4+) × (OH-)]/ NH3 ‫؞‬ Kb = (OH-)2/ NH3 (OH-)2 = Kb × (NH3) Before doing anything we have to calculate the volume of HCl required for complete neutralization, (M × V)NH3 = (M × V)HCl (50 x 0.1)NNH3 = (0.1 ×V)HCl  VHCl= 50 ml Before the equivalence point, after addition of 10 ml HCl: 1. Major species (before any reaction occur): NH3, H+, Cl- and H2O 2. The NH3 will react with H+ from the added HCl: NH3(aq) + H+(aq) → NH4+(aq) This reaction proceeds to completion as NH3 readily reacts with free proton. The stoichiometric calculations are then carried out using the known volume of 0.1 M HCl added. 3. After the reaction is complete, the major species will be NH3, NH4+, Cl- and H2O. The equilibrium can be represented as Acid (Ka) NH4+(aq) → NH3(aq) + H+(aq) + Cl- or Base (Kb) NH3(aq) + H2O(l) → NH4+(aq) + OH-(aq)  NH3 + HCl → NH4+ + H2O NH3 HCl Before 50.0 ml x 0.1 M = 10 ml x 0.1 M - Reaction = 5.0 mmol = 1.0 mmol ≡ 1 mmol NH4+ After = 4.0 mmol 0.00 NH3 + H2O ↔ NH4+ + OH- (0.1- X) X X Kb = [(NH4+) × (OH-)]/ NH3 ‫؞‬ Kb = (OH-)2/ NH3 (OH-)2 = Kb × (NH3) pOH = -log (OH-)  Titrations of weak bases with strong acids (cont.) At the equivalence point 1. By definition, the equivalence point occurs when all the original NH3 is converted to NH4+, the major species are NH4+, Cl- and H2O. 2. The dominant equilibrium will be the dissociation of the weak acid NH4+ for which Ka= Kw/Kb(for NH3). NH3 + HCl → NH4+ + H2O Before (50.0 ml x 0.1 M) 50 ml x 0.1 M NH3 HCl - Reaction = 5.0 mmol 5.0 mmol ≡ 0 mmol NH4+ After = 0.00 0.00 ≡ 5 mmol NH4+ NH4+ + H2O → NH4OH + H+ 5 mmol x x Ka = [NH4OH] [H+] = x2 and Ka = Kw = 1.0 × 10-14 = 5.6 x 10-10 [NH4+] 0.05 Kb 1.0 × 10-9 [H+]2 = Ka × 0.05 weak Acid-Strong Base Titration Curve (cont. IIl) Region 4. After the equivalence point, after adding 60.0 mL of 0.100 M NaOH. (Now calculate excess OH-) Total moles of OH- = -Moles of weak acid consumed = Moles of OH- remaining = - Beyond the equivalence point: 1. Excess HCl has been added and the major species are H+, NH4+, Cl- and H2O. 2. Although NH4+ will dissociate, it is a weak acid thus [H+] will be determined by the excess H+. If 60 ml HCl added, then: NH3 + HCl → NH4+ + H2O Before (50.0 ml x 0.1 M) 60 ml x 0.1 M NH3 HCl - Reaction = 5.0 mmol = 6.0 mmol ≡ 0 mmol NH4+ After = 0.00 0.00 ≡ 5 mmol NH4+ [H+] = 1 mmol H+ in excess / (110) volume solution in ml pH = - log [H+] = - log 0.009 = Titration curve of weak base with a strong acid The four Major Differences Between a Weak Acid- Strong Base Titration Curve and a Weak Base-Strong Acid Titration Curve 1. The initial pH is above 7.00. 2. A gradually decreasing portion of the curve, called the buffer region, appears before a steep fall to the equivalence point. 3. The pH at the equivalence point is less than 7.00. 4. Thereafter, the pH decreases slowly as excess strong acid is added.  Acid –Base indicators The most common acid-base indicators are complex molecules that are themselves weak acids or bases represented. They exhibit one color when the proton is attached to the molecule and a different color when the proton is absent. For example: Methyl red is a weak organic acid which can be used as an indicator in the pH range of 4.4 to 6.2. This implies that a solution of methyl red will be red if the pH is lower than 4.8 and yellow if it is above 6.2. On the other hand, if the pH of the solution is in this range (4.4< pH > 6.2), the colour will be an appropriate mixture of both the colours. - p- nitrophenol (weak acid), a commonly used indicator, is colorless in its HIn form and yellow in its In-, or basic form. - The undissociated form is colorless but the anion , which has a system of alternating single and double bond ( a conjugated system) is yellow. - Molecules having such conjugated system absorb light of longer wavelengths than comparable molecules in which no conjugated systems. Colorless Yellow Ph.Ph. in acidic medium Ph.Ph. in Basic medium Methyl orange in basic medium Methyl orange in acidic Determining the color change range of an indicator: - The dissociation of both acidic and basic indicators: (red) HIn + H2O H3O+ + In- (yellow) In- + H2O InH + OH- - The dissociation constant of the acidic indicator is Ka = [H3O+] [In-] so pH = pKa- log [HIn] [HIn] [In-] - Both forms [HIn (red), In- (yellow)] are present in solution and their relative concentration depends on pH. -At low pH (acidic medium), only red color exists (HIn), -at high pH (basic medium), yellow color exists (In-) -then and at intermediate pH the color is orange.  General Rule For most indicators, about one tenth (1/10) of the initial form of the indicator should be converted to the other form before a new colour is apparent so that the human eye can detect this colour change. We will assume, then, that in the titration of an acid with a base, the colour change will occur at a pH interval equal 2 pH units where pH = pKa± 1 Color change interval: HIn + H2O H3O+ + In- Ka = [H3O+] [In-] ‫؞‬ [H3O+] = Ka [HIn] [HIn] [In-] pH = pKa+ log [In-] [HIn] If [In-]=10 and [HIn] = 1 so the indicator is strongly dissociated ‫ ؞‬pH = pKa+ log 10 pH = pKa+ log 10 1 ‫ ؞‬pH = pKa+ 1 If [HIn] = 10 and [In-] =1 so the indicator is weakly dissociated ‫ ؞‬pH = pKa+ log 1 pH = pKa - log 10 10 ‫ ؞‬pH = pKa- 1 pH = pKa± 1  - Suppose pKa of HIn (weak acid) is 5. At low pH → [HIn]/[In-] is 10:1 red At high pH → [HIn]/[In-] is 1:10 yellow - So the minimum change in pH (∆ pH) required to cause a change in color (indicator range) from red to yellow is 2 units: Yellow : pHy = pKa + log 10/1 = 5+1 = 6 Red : pHr = pKa + log 1/10 = 5-1 = 4 ∆ pH = pHy – pHr = 6-4 = 2 - The indicator range in this example (4-6). - At an intermidiate pH (5), pKa = 5 because HIn is half neutralized [HIn] = [In-], the color is orange. Choice of Indicator for a certain titration When we choose an indicator for a titration, one should select an indicator which changes color at approximately the pH at the equivalence point of the titration. Choosing an indicator is easier when there is a large change in pH near the equivalence point as in case of strong acid-strong base titration, where the pH changes from 5.3 to 8.7 so any indicator working in the range from 5-9 can be used. While in case of weak acid-strong base titration, e.g. (HC2H3O2 against NaOH) the pH at the end point was 8.7. The choice is not that easy as we should use an indictor close to this value. As clear from the curves, of titration of a strong acid (to the left) and the titration of the weak acid (to the right) against NaOH, that : both phenolphthalein and methyl red can be used to determine the end point fort the strong acid titration. While for the weak acid only phenolphtalein can be used. Example: Calculate the change in pH that occurs when 0.01 mol solid NaOH is added to 1.0L of the buffered solution described in the previous example. Compare this pH change with that which occurs when 0.01 mol solid NaOH is added to 1.0 L of water. Since the added solid NaOH will completely dissociate, the major species in solution before any reaction occurs are HC2H3O2, Na+, C2H3O2-, OH-. Note that the solution contains a relatively large amount of very strong base which has great affinity for protons. OH- + HC2H3O2 H2O + C2H3O2- 56 Example (cont.) The best approach to this problem involves two distinct steps: 1. Assume that the reaction goes to completion and carry out the stoichiometric calculations then, 2. Carry out the equilibrium calculations. 1. The stoichiometry problem, HC2H3O2 + OH- C2H3O2- + H2O Note that 0.01 mol HC2H3O2 has been converted to 0.01 mol C2H3O2- by the added OH-. 57  Example (cont.) 2. The equilibrium problem, after the reaction between OH- and HC2H3O2 is complete, the major species are (HC2H3O2, weak acid) , (Na+, neither acid nor base), (C2H3O2-, base, conjugate base of acetic acid). This problem is now similar to the previous example, the only difference is that the addition of 0.01 mol of OH- has consumed some HC2H3O2, and produced some C2H3O2-, Thus, 58  Example (cont.) x = [H+] = 1.7x10-5 pH = 4.76  The change in pH produced by the addition of 0.01 mol OH- to this buffered solution is then 4.76 - 4.74 = + 0.02 pH unit 59  Example (cont.) Compare this with what happens when 0.01 mol solid NaOH is added to 1.0L water to give 0.01M NaOH. - In this case [OH-] = 0.01 M, thus - [H+] = Kw/[OH-] = 1.0x10-14/1.0x10-2 = 1.0x10-12 pH = 12.00  Thus the pH change is 12.00-7.00 = +5.00 pH units. Note how well the buffered solution resists a change in pOH as compared with pure water. 60  Henderson-Hasselbalch equation For the following acid dissociation reaction, HA H+ + A- This log form of the expression for Ka is called Henderson- Hasselbalch equation and is useful for calculating the pH of solutions when the ratio of [A-]/[HA] is known. 61  Henderson-Hasselbalch equation (cont.) For a particular buffering system (acid-conjugate base pair), all solutions that have the same ratio [A-]/[HA] will have the same pH. For example, a buffered solution containing 5.0M HC2H3O2 and 3.0M NaC2H3O2 will have the same pH as one containing 0.05M HC2H3O2 and 0.03M NaC2H3O2. This can be shown as follows: For (5.0M HC2H3O2 and 3.0M NaC2H3O2 ) , [A-]/[HA] = 3.0/5.0 = 0.6 For (0.05M HC2H3O2 and 0.03M NaC2H3O2 ), [A-]/[HA] =0.03/0.05 = 0.6 Therefore, pH = pKa + log (base/acid) = 4.74 + log (0.6) = 4.74 - 0.22 = 4.52 62  Example Calculate the pH of a solution containing 0.75 M lactic acid (HC3H5O3)(Ka=1.4x10-4) and 0.25M sodium lactate. HC3H5O3(aq) H+(aq) + C3H5O3-(aq) Applying Henderson- Hasselbalch equation. pH = pKa + log (salt/acid) pH = 3.85 + log (0.25/0.75) = 3.38 63 64 65  Experiment 4 Acid- Base Titration Determination of the concentration of sodium Carbonate and sodium Bicarbonate in mixture Procedure: 1) Washing the burette with distilled water then with HCl. 2) Washing the pipette with distilled water then with mixture. 3) Washing the flask with distilled water. 1) Titrate 10 ml of the mixture with standard HCl in the presence of 2 drops of ph.ph. (Pink------faint pink) 2) Titrate 10 ml of the mixture with standard HCl using MO as indicator. Repeat each step at least three times. (Yellow----orange)