We remember that molarity (M) times volume (V) is equal to moles so this relationship is the definition of the equivalence point.

Transcription

1 Titrations Titration - a titration is defined as the determination of the amount of an unknown reagent (analyte) through the use of a known amount of another reagent (titrant) in an essentially irreversible reaction. Acid-Base titrations In this section we will discuss titrations of acids and bases. The unknown acid or base can be weak or strong, but the known acid or base that we use as the titrant must be strong so the reaction will be irreversible. The following is the general reaction between a mono protic acid and the strong base hydroxide. HA(aq) + OH - (aq) > H2O(aq) + A - (aq) As we add hydroxide to the acid solution we will use up the acid (HA) and make the conjugate base (A - ). When we have added equal amounts of base to the acid we say we are at the equivalence point of the titration. At this point the moles of added base are equal to the initial moles of the acid. With this in mind the following relationship between the volume and concentration of the acid and base can be written. Macid x Vacid = Mbase x Vbase We remember that molarity (M) times volume (V) is equal to moles so this relationship is the definition of the equivalence point. Exercise- using the equivalence point relationship answer the following questions. 1. How many ml of 0.2 M NaOH would be required to react completely with 25 ml of 0.5 M HCl. 2. How many ml of 0.5 M NaOH would be required to react completely with 35 ml of 0.25 M acetic acid. 3. How many ml of 0.8 M NaOH would be required to react completely with 15 ml of 1.5 M nitrous acid. 48

2 Determination of species present as base is added and calculation of ph It is important to be able to calculate the theoretical ph as we do a titration. To do this we will use the following stoichiometry reaction for an acid, followed by the use of the Henderson hasslebach equation. The use of the following stoichiometry equation will allow us to determine the amount of the conjugate acid and base. HA + OH > H2O + A - We put in our values and treat this as a limiting reactant problem in which the base is limiting ml of 1M HNO2 was titrated with 0.2 M NaOH. How many moles of acid and conjugate base will be present after the addition of 5 ml of sodium hydroxide? Answer- we first set up the reaction. HA + OH > H2O + A - Initial (25 ml)(1m) (5 ml)(0.2) 0 25 mmol 1mmol Change - 1mmol - 1mmol +1mmol Finish 24 mmol 0 1mmol 24 mmol HNO2 1 mmol NO ml of 2 M HC2H3O2 was titrated with 0.5 M NaOH. How many moles of acid and conjugate base will be present after the addition of 10 ml of sodium hydroxide? Answer- we first set up the reaction. HA + OH > H2O + A - Initial (15 ml)(2 M) (10 ml)(0.5) 0 30 mmol 5 mmol Change - 5mmol - 5mmol +5mmol Finish 25 mmol 0 5mmol 25 mmol HC2H3O2 5 mmol C2H3O2-49

3 Exercises ml of 2 M NH4 + was titrated with 0.15 M NaOH. How many moles of acid and conjugate base will be present after the addition of 50 ml of sodium hydroxide? 2. 2 ml of 0.3 M HC3H5O2 was titrated with 0.5 M NaOH. How many moles of acid and conjugate base will be present after the addition of 0.13 ml of sodium hydroxide? ml of 0.25 M HCOOH was titrated with 0.18 M NaOH. How many moles of acid and conjugate base will be present after the addition of 16 ml of sodium hydroxide? ml of 0.1 M HCN was titrated with 0.23 M NaOH. How many moles of acid and conjugate base will be present after the addition of 9 ml of sodium hydroxide. 50

4 Calculation of ph Initial point To calculate the initial ph we use an ICE table and Ka to solve for the hydronium concentration which can then be used to get the ph. ( review acid handout) Excercises ( use the Ka values from your book.) 1. What is the ph of a 0.2M solution of Acetic acid? 2. What is the ph of a.002m solution of HNO2? After addition of base To calculate the ph we use the Henderson-Hasselbach equation. ph = pka + log base acid Once we have determined the moles of acid and base( using method of first part of handout) we can then plug them into the equation and solve for the ph. What is the ph of the solution from example 2 in the previous section. 51

5 Answer We found that there are 25 mmol of HC2H3O2 and 5 mmol of C2H3O2 -. These are the values that will be plugged into the Henderson hasslebach equation. ph = pka + log base acid = log 5 25 = Exercise 1. What is the ph of the solution when 25 ml of 0.1 M NaOH is added to 50 ml of 0.3M H3PO4. 2. What is the ph of the solution when 15 ml of 0.25 M NaOH is added to 55 ml of 0.4M HNO2. ph at the equivalence point The ph at the equivalence point is a two part problem. The first part of the problem involves finding the concentration of the conjugate base, and the second part of the problem is to use the base equilibrium to find OH -. The Kw relationship can then be used to solve for H3O + which can be used to solve for ph. What is the ph at the equivalence point for 50ml of a 0.1M solution of acetic acid, titrated with 0.2M NaOH. Answer The first thing we must determine is the amount of base required to reach the equivalence point. Macid x Vacid = Mbase x Vbase V base = (V acid )(M acid ) (M base ) = (50 ml)(0.1m) 0.2M = 25 ml Now we must calculate the molarity of the conjugate base to use in the base equilibrium. First we use our stoichiometry equation. 52

6 HAc + OH > H2O + Ac - Initial (50 ml)(0.1m) (25 ml)(0.2) 0 5 mmol 5mmol Change - 5 mmol - 5 mmol 5 mmol Finish mmol This shows us that we have 5mmol of base, which we can divide by the total volume ( ml s of initial acid + ml of base required to get to end point) in ml s to get the concentration. Concentration = 5 mmole 75 ml = 0.67M We now use the base equilibrium to determine hydroxide concentration. Ac - + H2O > OH - + HAc K b = Kw Ka We use an ice table with the initial concentration of the base determined above in a Kb problem. ( review the base handout) ml of a weak acid required 20 ml of 0.1M NaOH to reach the equivalence point what is the ph at the equivalence point. Ka = 5.7 x 10-5 Answer We first determine the moles of the conjugate base. As we saw before assuming it is a monoprotic acid the moles of base required to reach the endpoint will equal the moles of the conjugate base at the equivalence point. Moles of conjugate base = 20 ml x 0.1M = 2 mmoles Since we want the ph at the equivalence point we will calculate the concentration of the base. concentration of conjugate base = 2 mmoles 65 ml = 0.03M This will be the initial concentration for the Kb ice table to determine the OH - concentration. 53

7 B - (aq) + H2O(l) < > HB(aq) + OH - (aq) Kb = 1.75 x Initial 0.03M 0 0 change -x +x +x equilibrium 0.03M - x x x [BH] [ OH - ] [B - ] = K b = [x] [ x ] [0.03] = 1.75 x x 2 = 1.75 x x 0.03 x = 2.3 x 10-6 = [OH - ] poh = 5.64 ph = 14 - poh = = 8.36 Exercise 1. Acetylsalicylic acid has a Ka of 3.0 x In a titration 35 ml of 0.25 M NaOH was required to reach the endpoint with 100ml acid what is the ph at the endpoint. 2. lactic acid has a Ka of 1.4 x In a titration 18 ml of M NaOH was required to reach the endpoint with 100ml of acid what is the ph at the endpoint. 54

8 Titration curve exercise What is the ph at the points below for the titration of 50 ml of 0.20 M Iodic acid (HIO3) with 0.1M NaOH. Ka = 1.2 x 10-5 Initial 10ml NaOH 50 ml NaOH 75 ml NaOH Equivalence point 55

9 Buffers Buffers are solutions of a weak acid and its corresponding conjugate base. The following are some examples of buffers. HC2H3O2/ NaC2H3O2 H2CO3/ NaHCO3 NaHCO3 / Na2CO3 NaH2PO4 / Na2HPO4 Acetic acid / sodium acetate Carbonic acid / sodium bicarbonate Sodium bicarbonate / sodium carbonate sodium dihydrogen phosphate / sodium hydrogen phosphate. Exercise Determine if the following acid base pair is a buffer. H3O + / H2O Na2HPO4 / Na3PO4 H2S / Na2S NH4Cl / NH3 ph of a buffer solution To calculate the ph of a buffer solution we use the Henderson hasslebach equation. ph = pka + Log Base Acid If we want to determine ph we need to know the pka and the base and acid concentrations. In a buffer made from acetic acid and sodium acetate ( pka for Hac =4.74 ) the acid concentration was 0.30 M and the base concentration was 0.60M what is the ph? Answer We have every thing we need to use the Henderson hasslebach equation. 56

10 ph = pka + Log Base Acid = log = 5.0 The concentrations do not need to be used, if we know how many moles of acid, and base are in the solution as the following example shows. 25 ml of 0.1 M sodium acetate was combined with 30 ml of 0.2 M acetic acid what is the ph of the resulting solution. Answer As in any problem in which we are to determine the ph of a solution of a weak acid and its conjugate base, we will use the HH eqn. We know the pka form the previous problem, and we will determine the moles of acid and base from the given data. mmole of acid = ml of acid x M acid = 25 ml x 0.10 M = 2.5 mmole mmole of base = ml of base x M base 30 ml x 0.20 M = 6.0 mmole Excercises ph = pka + Log Base Acid = log 6.0mmole 2.5 mmole = 5.1 Calculate the ph of the following buffer solutions M NaHCO3 / 0.4M Na2CO M HC2H3O2 / 0.4M C2H3O ml 1.2 M H3PO4 / 45 ml 1.0M NaH2PO ml.50 M NaHC2O4 / 35 ml 0.3 M Na2C2O4 57

11 Calculation of ph after addition of a strong acid or base to a buffer. By definition, buffers are solutions that resist a change in ph upon addition of a strong acid or base. This takes place because the strong acid or base is transformed into a weak acid or bases. This is shown in the reactions below. Reaction of strong added base with weak acid in buffer to make weak base. 1. HA(aq) + OH - (aq) > A - (aq) + H2O(aq) Reaction of strong added acid with weak base to make weak acid. 2. A - (aq) + H3O + (aq) > HA(aq) + H2O(aq) ph of a buffer solution after the addition of a strong acid or base. The calculation of the ph of a buffer involves two steps. The first step is the determination of the acid, conjugate base amounts after the addition of the strong acid or base. This involves stoichiometry, using either equation 1 or 2 above. To do this we will set up a table as in the following example. What is the amount of the acid, and conjugate base after addition of 10 ml of 1M HCl ( 10 mmole), to a buffer that is 35 mmole of weak acid HA, and 45 mmole weak base A -. Step 1 Write out the appropriate equilibria, which in this case is equilibria 2. A - (aq) + H3O + (aq) > HA(aq) + H2O(aq) Step 2 Make the following table underneath the equation A - (aq) + H3O + (aq) > HA(aq) + H2O(aq) initial Change Final 58

12 Step 3 Fill in the values from the givens in the problem. A - (aq) + H3O + (aq) > HA(aq) + H2O(aq) initial 45 mmole 10 mmole 35 mmole Change Final Step 4 Subtract the amount of acid from the weak base and the acid, and add the amount of acid to the weak acid. A - (aq) + H3O + (aq) > HA(aq) + H2O(aq) initial 45 mmole 10 mmole 35 mmole Change - 10 mmole - 10 mmole +10 mmole Final Step 5 Calculate the final amount for all the species in solution. A - (aq) + H3O + (aq) > HA(aq) + H2O(aq) initial 45 mmole 10 mmole 35 mmole Change - 10 mmole - 10 mmole +10 mmole Final 35 mmole 0 mmole 45 mmole We can then plug these results into the HH equation with the appropriate pka to determine the ph. The same procedure is done with the addition of strong bases, but in this case we use equation 1. 59

13 What is the final concentration of weak acid and conjugate base, when 4 mmole of OH - is added to a buffer that contains 30 mmole HA and 20 mmole A -. Answer HA(aq) + OH - (aq) > A - (aq) + H2O(aq) initial 30 mmole 4 mmole 20 mmole Change - 4 mmole - 4 mmole + 4 mmole Final 26 mmole 0 mmole 24 mmole What is the ph of a acetate buffer, which contains 25 mmole acetic acid, and 30 mmole of sodium acetate, after the addition of 15 mmole HCl? Answer C2H3O2 - (aq) + H3O + (aq) > HC2H3O2(aq) + H2O(l) pka = 4.74 initial 30 mmole 15 mmole 25 mmole Change -15 mmole -15 mmole +15 mmole Final 15 mmole 0 mmole 40 mmole ph = pka + Log Base Acid = log = 4.31 What is the ph of a buffer made from 25 ml 0.2M NaH2PO4 and 35 ml 0.1M Na2HPO4, after addition of 10 ml of 0.1M NaOH? 60

14 Answer H2PO4 - (aq) + OH - (aq) > HPO4 2- (aq) + H2O(l) pka = 7.21 The first thing to do is calculate the initial mmols of each species. H2PO4 - = 25 ml x 0.2M = 5 mmole OH - = 10 ml x 0.1M = 1 mmole HPO4 2- = 35 ml x 0.1M = 3.5 mmole H2PO4 - (aq) + OH - (aq) > HPO4 2- (aq) + H2O(l) initial 5 mmole 1 mmole 3.5 mmole Change - 1 mmole - 1 mmole + 1 mmole Final 4 mmole 0 mmole 4.5 mmole Exercise ph = pka + Log Base Acid = log = A phosphate buffer was made by combining 20 ml of 1M NaH2PO4 and 15 ml of 2M Na2HPO4, then 3 mmole of HCl was added to the buffer, what is the ph? 61

15 2. An enzymatic reaction was performed in 50 ml of a carbonate buffer, made from equal volumes of 0.1M sodium carbonate and sodium hydrogen carbonate. The reaction produced 0.6 mmole of hydroxide upon completion, what is the ph. 3. A oxalate buffer was found to contain 3.4 mmole NaHC2O4 and 2.6 mmole Na2C2O4. If 3 ml of 0.5 M sodium hydroxide was added to the buffer what is the ph? 62

16 Preparation of buffers A buffer is a solution of a weak acid and its conjugate base, in which we find the ph dependent on the ratio of the acid to the base. Most buffer problems will ask us to prepare a certain amount of buffer at a given ph. To prepare the buffer we will divide the process into 3 steps and discuss each step individually. Step 1 Determining the moles of acid and Base. How many moles of acid and base would be required to prepare a 2.00L solution of ph = M carbonate buffer. pka1 = 6.4 pka2 = Answer First we determine the acid by comparing the ph of the buffer with the pka values of the available acids. In this case the ph is closest to the second pka so we will use HCO3 - as our acid and CO3 2- as the base. A = HCO3 - B = CO3 2-1 Write out the HH eqn. ph = pka + log B A 2 Plug in the pka value and the ph. 9.8 = log B A 3. Solve for the log of the ratio of B to A. log B A = ph - pka = = Take the inverse log of both sides to get the ratio. B A = = Solve for B in terms of A B = 0.63 A 63

17 We will now use the volume of the buffer and its molarity to determine the total moles in the buffer solution, which is a combination of the acid and the base moles. B + A = M buffer V buffer = (1M)(2L) = 2 moles We plug in our value of B determined previously and solve for A A + A = 2 moles 1.63 A = 2 moles A = 2 moles 1.63 = 1.22 moles Since A + B = 2 B = 2 - A = = 0.78 A= 1.22 moles B = 0.78 moles Exercise (using the steps above) 1. Calculate the moles of acid and base needed to make 500 ml of a M phosphate buffer at ph How many moles of acid and base would be required to make 2.2 l of a 0.56M carbonate buffer at ph 9.7? 64

18 3. How many moles of acid and base would be required to make 2.2 l of a 0.56M carbonate buffer at ph 9.7? Preparation of a buffer given moles of acid and base. In this section we are going to determine the volume of soution or grams of solid required to give the moles of acid or base from our previous calculations. We will also investigate 1. 1 L of a buffer was to be prepared from 1M Na2HPO4 and 1M NaOH, if 0.5 moles of acid, and 0.2 moles of conjugate base was required, calculate the volume of both solutions required to give those amounts. Answer Since we only have one phosphate source, all phosphate species must come from the Na2HPO4. moles of Na2HPO4 = 0.5 moles + 0.2moles = 0.7 moles moles of NaOH = moles of base = 0.2 moles. Volume of the acid and base Volume of 1M NaOH 0.2 moles NaOH x 1 1 mole L = 0.2L = 200 ml Volume 1M Na2HPO4 0.7 moles Na 2 HPO 4 x 1 1 mole L = 0.7L = 700 ml 65

19 To prepare a carbonate buffer we required 0.10 moles of acid and 0.23 moles of base. If we have solid Na2CO3 and 1M HCl, how many grams of the sodium cabonate and milliliters of acid will be required. Answer To determine the grams of carbonate we must find the molecular weight of the compound. The weight is found to be and we do the following calculation to find the grams. Since the Na2CO3 is the only source of both the acid and the base the total moles needed will be 0.33 mole. grams Na2CO3 = 0.33 moles x grams/mole = grams volume HCl = 0.23 moles/ (1 mole/l) = 0.23 L = 230 ml Exercise 1. in the preparation of an acetate bufer 1.2 moles of acid was required and 0.2 moles of acid was required. If we have 1M acetic acid and solid sodium hydoxide what volume of the acid and mass of the hydroxide would be required. 2. An oxalate buffer required 0.75 moles of H2C2O4 and 0.13 moles of HC2O4-. If both components are available as 1M solutions what volume of each solution is required. 3. What are the volumes of 0.45M KHCO3 and 0.50 M K2CO3 required to prepare 80 ml of a buffer that contains 0.4mmol acid and 0.2 mmol base. 66

20 3. How many grams of sodium acetate ( MW = 81.99), and what volume of 0.50 M HCl or 0.50 M NaOH, would be required to prepare a buffer containing 6 mmol acid and 8 mmol base. Combined buffer problems Describe how to prepare 1L of a 0.1M phosphate buffer at ph 6.4. You are to use 1M KH2PO4, ( MW = 136.1g/mole ) and either 1M HCl or 1M NaOH. Answer The following steps apply to all buffer problems in which you are given a ph and an amount of buffer. We will combine all of the previous procedures. Step 1 Pick an acid(use tables from the text), by finding the acid that is about +/ - 1 pka unit different from the desired ph. In the case of a phosphate buffer the following acids and corresponding pkas are available. H3PO4 pka1 = 2.5 H2PO4 - pka2 = 6.8 HPO4 2- pka3 =10.9 In our example since the ph is 6.4 we will use pka2, and our acid will be H2PO4 therefore, the conjugate base will be HPO4 2-. Step 2 Once we have our pka and the conjugate acid base pair, we will use the Henderson hasslebach (HH ) equation to determine the acid(a) base (B) ratio. 67

21 Procedure 1 Write out the HH eqn. ph = pka + log B A 2 Plug in the pka value and the ph. 6.4 = log B A 3. Solve for the log of B to A. log B A = ph - pka = = Take the inverse log of both sides. B A = = Solve for B B = 0.4 A Step 3 We will now use the volume of the buffer and its molarity to determine the total moles which is a combination of the acid and the base moles. B + A = M buffer V buffer = (0.1M)(1L) = 0.1 moles We plug in our value of B determined previously and solve for A. 0.4 A + A = 0.1 moles 1.4 A = 0.1 moles A = 0.1 moles 1.4 =.07 Since A + B = 0.1 B = A = =

22 Preparation of the buffer Now that we have the concentrations of both the acid and the base, we can go back to the problem to determine what we have to make our buffer from. In this case all we have is KH2PO4, and either HCl or NaOH. In a case like this, since there is only one source of phosphate (KH2PO4), the total moles of phosphate will be equal to our initial amount of KH2PO4. We know that we need the H2PO4 - as our acid so the conjugate base needs to be made from the acid by the addition of NaOH. The amount of base added will be equal to the amount of conjugate base needed. The following reaction describes the process. KH2PO4(aq) + NaOH(aq) > NaKHPO4(aq) + H2O(l) initial 0.1 moles 0.03 moles 0 change moles moles moles end 0.07 moles moles This give us the correct amount of the acid and its conjugate base for the desired ph. Now to prepare the buffer we must determine the grams of solid KH2PO4 and the volume of aqueous 1M NaOH. Mass of solid KH2PO4 To determine the mass of the solid compound we multiply the desired moles by the MW of the compound moles KH 2 PO 4 x g mole = 13.6 g KH 2 PO 4 We use O.100 moles, because KH2PO4 is our only phosphate source. Volume of 1M NaOH To determine the volume of aqueous compounds we divide the desired moles by the molarity of the solution moles NaOH x 1 1 mole L = 0.03L = 30 ml Final preparation Since we want one liter of buffer we will use a 1L volumetric flask. To this flask we will add 13.6 g of KH2PO4 and 30 ml of 1M NaOH., and dilute to the mark. 69

23 Combined buffer problems 1. Prepare 1L of a 0.01 acetate buffer at ph 5.0. You have 0.1M acetic acid and 0.2 M HCl or 0.2 M NaOH. 2. Enzymatic reactions are most efficient at a limited range of ph. To run an enzyme reaction, 3L of 0.35M phosphate buffer at ph 7.1 was required. In the lab a solution of 0.8M NaH2PO4 was found. There is also solid NaOH (MW= 40) and 1M HCl that can be used in the preparations of the buffer. Describe how to make this buffer. 3. In the preparation of a Ca(OH)2 solution it was desired to maintain the ph at 9.3. If 1 L of 0.6M buffer is required for this experiment, and the following acids and bases are available, describe how you would prepare this buffer. Solid NaHCO3 pka1 = Solid NH4Cl pka = M NaOH 1M HCl 70