Chemistry 30 Acid Base Neutralization and Titration

Transcription

1 Chemistry 0 Acid Base Neutralization and Titration This unit will cover a number of interconnected topics: 1. Acid base titration to find concentration curves calculate the change in as an acid is added to a base or a base to an acid identifying unknown acids and bases from the shape of the curve choosing indicators 2. Writing acid base neutralization reactions Molecular (complete neutralization) Steps during the titration. Indicators Writing reactions (including endpoint) Choosing indicators Determining the of a solution from indicator colour Adding indicator colours together Titration - The gradual addition of one solution to another until the reaction is complete (as indicated by an indicator Endpoint when the indicator dictates that the reaction is complete Indicator may be a chemical species, a change in temperature, changes in conductivity or changes in - acid/base indicators are weak acid/weak conjugate base pairs that change colour when they donate or accept a proton - an acid/base neutralization reaction releases heat so as the reaction progresses, the temperature will continue to rise, spiking at the equivalence point Equivalence point when the reaction is complete (acid base) Buret - graduated cylinder with a stopcock at the bottom. The graduations start at 0 at the top, so it measures how much is ADDED to the erlenmeyer flask. - - fill with either acid or base (usually of known concentration) called the TITRANT Erlenmeyer flask - fill with the substance being titrated (usually unknown concentration), 10 to 25 m of the substance measured with a graduated cylinder or pipette. - add the indicator to the Erlenmeyer flask 2 drops of indicator, a meter, thermometer, etc. - swirl the flask as you add the titrant (or use a magnetic stirrer) - add titrant until the endpoint is reached (a permanent colour change, a spike in temperature, a steep increase in ) Always read the bottom of the meniscus and to one digit past the last line m Chemistry 0 Unit VI Notes page 1

2 Titration Curves - A titration curve is a plot of the of the mixture in the Erlenmeyer flask vs. the volume of titrant added. - You must be able to: - Identify the acid/base being titrated from the shape (or draw the shape of the curve given the acid/base combination) - Choose indicators for the titration given the curve - Identify the equivalence point - Write reactions for the titration - Calculate concentration (stoichiometry) Part I Shapes - The shape of a titration curve is determined by the concentration and strength of both the acid or base being titrated and the titrant. The titration of NaOH(aq) with HCl(aq) (strong base with strong acid) This shape is characteristic of the titration of a strong base with a strong acid (the titrant is always written 2 nd, after the with ) - The strong base causes the to start high and have a buffering region (flat part) at the beginning of the titration where the changes little as the acid is added. The initial can be calculated using poh log[oh - ] then poh 7 - When the reaction is close to complete (nearing the equivalence point) there will be a large drop in. - The midpoint of the drop is the equivalence point, when Volume of HCl added (m) [acid] [base] (choose your indicator to change colour at this, and use this volume for stoichiometry) Since this curve represents the titration of a strong base (in the Erlenmeyer flask) with a strong acid (the titrant), the net ionic equation would be: H O (aq) OH - (aq) 2 H 2 O(l) (all strong acids are ionized into H O (aq), and all strong bases are salts with OH - (aq)) - Before the equivalence point, there is excess base in the Erlenmeyer (the is high) - After the equivalence point, there is excess acid in the Erlenmeyer (the is low) - At equivalence, equal amounts of acid and base are present (the net ionic equation as written has stoichiometric quantities of both acid and base the reaction is complete). Since the only product is water, the is based on the of water ([H O (aq)] 1.00 x 10-7 mol/, and 7). This is true of all titrations of strong acid/strong base since the only product is water. The titration of HCl(aq) with NaOH(aq) (strong base with strong acid) Similarily, a titration curve for a strong acid with a strong base would look like: 7 Volume of NaOH added (m) - The starts low since it is a strong acid ( -log [H O (aq)]), and it follows the same shape (but in the opposite direction) as the titration of the strong base with strong acid. - The midpoint of the steep part is the equivalence point, the will be 7 at equivalence (as with strong base with strong acid) - The final will be very close to the of the base that is being added The net ionic equation for this titration is the same as for the titration of a strong base with a strong acid, H O (aq) OH - (aq) H 2 O(l) Chemistry 0 Unit VI Notes page 2

3 When a weak acid or base is being titrated, a couple of distinct features are noted: - The buffering region when only the weak acid or base is present is not as flat as if it were a strong acid or base - The steep portion as equivalence is reached is not as steep - The at equivalence is relfective of the contents at equivalence The titration of CH COOH(aq) with NaOH(aq) (weak acid/strong base) >7 - The weak acid s initial (at the same concentration) will not be as low as a strong acid s - The buffering region at the beginning is not as flat - The steep region through the equivalence point is not as steep The net ionic equation is: CH COOH(aq) OH - (aq) CH COO - (aq) H 2 O(l) (the weak acid is not ionized in this reaction) Volume of NaOH added (m) - Since the products are water and a weak base (CH COO - (aq)), the will reflect the presence of the weak base and be greater than 7 (this is true of all weak acid (WA) strong base (SB) titrations) The titration of NH (aq) with HCl(aq) (weak acid/strong base) <7 - The weak base s initial (at the same concentration) will not be as high as a strong base s - The buffering region at the beginning is not as flat - The steep region through the equivalence point is not as steep The net ionic equation is: NH (aq) H O (aq) NH 4 (aq) H 2 O(l) (the weak base is not ionized in this reaction) Volume of HCl added (m) - Since the products are water and a weak acid (NH 4 (aq)), the will reflect the presence of the weak acid and be lower than 7 (this is true of all weak base (WB) strong acid (SA) titrations) Polyprotic acid and base titrations are different still since it is really more than one titration taking place. Each proton transferred represents a separate curve and a separate equivalence point. The titration of a diprotic acid with a strong base will have two equivalence points, a triprotic will have three. The titration of H 2 CO (aq) with NaOH(aq) (weak diprotic acid with strong base) Volume of HCl added (m) - Since H 2 CO (aq) is a weak acid, its initial is higher than a strong acid of the same concentration and the initial buffering region is not as flat. - Since H 2 CO (aq) is a diprotic acid, there are two equivalence points, represented by the following net ionic equations: 1. H 2 CO (aq) OH - (aq) HCO - (aq) H 2 O(l) 2. HCO - (aq) OH - (aq) CO 2- (aq) H 2 O(l) The first equivalence point is the complete transfer of the first proton The second equivalence point is the complete transfer of the second proton) Chemistry 0 Unit VI Notes page

4 - The s at the equivalence points are representative of the products at equivalence. At the first equivalence point in the titration of a diprotic substance, the products will always be water and an amphoteric substance. The will be determined by the of that substance (is it a better base or a better acid?) - The at the second equivalence point will always be higher than that of the first equivalence point, and since the second acid in a diprotic substance is always weaker than the first, the shape of the curve will be characteristic of a weaker acid (a less flat buffering region and a less steep equivalence region) - You can always identify how many protons are present in the polyprotic acid by how many steep sections there are on the titration curve (this curve has two steep sections, so it is a diprotic, and it is an acid since the starts low) The titration of Na 2 CO (aq) with HCl(aq) (weak diprotic base with strong acid) Volume of HCl added (m) - Since Na 2 CO (aq) is a weak base, its initial is lower than a strong base of the same concentration and the initial buffering region is not as flat. - Since Na 2 CO (aq) is a diprotic base (or dibasic), there are two equivalence points, represented by the following net ionic equations: 1. CO 2- (aq) H O (aq) HCO - (aq) H 2 O(l) 2. HCO - (aq) H O (aq) H 2 CO (aq) H 2 O(l) The first equivalence point is the complete transfer of the first proton The second equivalence point is the complete transfer of the second proton) The titration of 10 mol/ H 2SO 4(aq) with NaOH(aq) Volume of HCl added (m) EXCEPTION! These shapes are appropriate for all titration curves except the titration of 10 mol/ (or more concentrated) H 2 SO 4 (aq) with NaOH(aq). Since this is diprotic, we would expect to have two clear equivalence points. BUT: The first and second proton are SOOO close in strength and since the first proton is strong, at 10 mol/ concentrations or higher, the first equivalence point will not be noticeable, and will appear strong. The second one will be seen, and will demonstrate the slightly weak nature of that proton: Part II Choosing Indicators - Indicators are weak acid/base conjugates that change colour when they accept or donate a proton - To be effective, the indicator should change colour (accept/donate a proton) as close to the equivalence point as possible (immediately after all of the reactant in the beaker is consumed) *** The indicator MUST be weaker than the acid or base whose equivalence point it indicates so that it will react after the acid or base Choose an indicator that: - Is ½ way between the acid/base being titrated and the next strongest acid/base in the Erlenmeyer flask (don t forget water) (use p.11) - Donates a proton at the appropriate at equivalence (use p.10 and p.11). - Diprotic acid/base titrations may have an indicator for the first equivalence, the second or both. Chemistry 0 Unit VI Notes page 4

5 ex: Choose an indicator for the titration of HCl(aq) with NaOH(aq). Since an acid is being titrated (in the Erlenmeyer flask), use the acid side of p.11 to choose your indicator. ist the species in the beaker right now, and identify them as A, B, A/B, s (spectator) and the SA, SB H O (aq) Cl - (aq) H 2 O(l) Na (aq) OH - (aq) (titrant) SA s A/B s SB The H O (aq) will be the SA and the titrant, OH - (aq) will be the SB. When they finish reacting, the equivalence point is reached. The indicator MUST be the next strongest acid in order to properly indicate the equivalence point. Since the only other acid in the beaker is H 2 O(l), the indicator should be between these two acids ideally, ½ way. Since the Ka of H O (aq) is 1 and the Ka of water is 1 x 10-14, halfway is 10-7, something like HBb(aq) (bromothymol blue). Since we know that the equivalence point is 7.0 (for the titration of strong acid and strong base), looking at page 10, bromothymol blue, with a Ka of 10-8 changes colour at a between 6.0 and 7.6, which encompasses the equivalence for this titration. At 7.0, bromothymol blue will be in the middle of its colour change from yellow to blue (the acid form is yellow as shown on p.10 and blue in the base form) so the endpoint (colour change) will be to green. ex: Write the equations for the reactions during the titration using bromothymol blue as an indicator. - As in the previous unit, to write reactions, list all of the species present. Since bromothymol blue is in the Erlenmeyer flask, when placed in the acid, it will be in its acid form (HBb(aq) as shown on p.10) - In titration reactions, the titrant will AWAYS appear in every reaction since it is always one of the strongest. Write reactions as in the previous unit the SA with the SB. H O (aq) Cl - (aq) H 2 O(l) HBb(aq) Na (aq) OH - (aq) (titrant) SA s A/B A s SB Put a box around the titrant to remind yourself that it will appear in every reaction. 1. H O (aq) OH - (aq) 2 H 2 O(l) This is the first reaction since H O (aq) is the SA and OH - (aq) is the SB. When this reaction is complete, the first equivalence point is reached. Now that the H O (aq) is used up, cross it off the list of species present. Water is produced, so it would be added to the list of species (except it is already there). The next substance that will react is the next strongest acid. Bromothymol blue is the next strongest acid. H O (aq) Cl - (aq) H 2 O(l) HBb(aq) Na (aq) OH - (aq) (titrant) SA s A/B SA s SB 2. HBb(aq) OH - (aq) H 2 O(l) Bb - (aq) Yellow blue Indicator reactions are written just like all other acid base reactions, but the colour change is included below the indicator. This reaction represents the enpoint, and is a good choice of indicator because it occurs right after the equivalence point. Chemistry 0 Unit VI Notes page 5

6 Now that the HBb(aq) is used up, cross it off the list of species present. Water is produced, so it would be added to the list of species (except it is already there). The next substance that will react is the next strongest acid. The only acid left is water. This reaction may be written, but is often omitted since there is really no reaction. H O (aq) Cl - (aq) H 2 O(l) HBb(aq) Na (aq) OH - (aq) (titrant) SA s A/B SA s SB. H 2 O(l) OH - (aq) H 2 O(l) OH - (aq) If this reaction is written, use equilibrium arrows since there is no change in the species. This represents the reaction that occurs in the last part of the titration curve.. H 2O(l) OH - (aq) H 2O(l) OH - (aq) 2. HBb(aq) OH - (aq) H 2O(l) Bb - (aq) (the indicator reaction which occurs right Yellow blue after equivalence) 1. H O (aq) OH - (aq) 2 H 2O(l) (the reaction until equivalence) Volume of NaOH added (m) ex: Choose an indicator and write reactions representing the steps in the titration of HOCl(aq) with NaOH(aq). Sketch the titration curve. - Since we don t know what the will be at equivalence (just that it s above 7), use p.11 to choose an indicator after listing all of the species present and writing the reactions. There will be two reactions, one for HOCl(aq) and one for H 2 O. HOCl(aq) H 2 O(l) Na (aq) OH - (aq) (titrant) A A/B s SB 1. HOCl(aq) OH - (aq) OCl - (aq) H 2 O(l) 2. H 2 O(l) OH - (aq) H 2 O(l) OH - (aq) - The indicator should be chosen ½ way between the two acids, HOCl(aq) and H 2 O(l). Using p. 11, HOCl(aq) has a Ka of 4.0 x 10-8, and water is 1.0 x Half way is On p. 11, there is only HIc(aq) to choose. On p. 10, we could also choose HTh(aq) (Ka ) or Hay(aq) (Ka ) (Any of these choices is as good as the other). - When you ve chosen the indicator you want, write the reaction where it would occur, which is inbetween the two reactions you already have (since it is a weaker acid than HOCl, but a stronger acid than water). So overall, Titration of HOCl(aq) with NaOH(aq) 1. HOCl(aq) OH - (aq) OCl - (aq) H 2 O(l) HTh(aq) OH - (aq) Th - (aq) H 2 O(l) Colourless blue PH >7 Equivalence point 2. H 2 O(l) OH - (aq) H 2 O(l) OH - (aq) Volume of NaOH(aq) added (m) Chemistry 0 Unit VI Notes page 6

7 To choose indicators for a titration of a base with an acid, just use the other side of p.11 since you are starting with bases). Ex: Write reactions for the titration of Na 2 CO (aq) with HCl(aq). Choose indicators for each of the equivalence points and write the reactions. Include the overall endpoint. We will write the reactions first, then choose indicators to fit between. (you can do both at once) - Since Na 2 CO (aq) is diprotic (dibasic), there will be two reactions during this titration. ist all of the species first, and write a reaction between the SA and SB. Na(aq) CO 2- (aq) H 2 O(l) Cl - (aq) H O (aq) (titrant) s SB A/B s SA Equiv. Point#1 CO 2- (aq) H O (aq) HCO - (aq) H 2 O(l) Cross out the CO 2- (aq) since it is used up, and add HCO - (aq) to the list since it is produced. Na(aq) CO 2- (aq) H 2 O(l) HCO - (aq) Cl - (aq) H O (aq) s SB A/B A/SB s SA - The next strongest base is HCO - (aq). So a reaction is written using the titrant and the SB. When the HCO - (aq) is used up, this marks the second equivalence point. Equiv. Point#2 HCO - (aq) H O (aq) H 2 CO (aq) H 2 O(l) Since H 2CO (aq) is produced, bubbles should be observed since H 2CO (aq) decomposes into water and carbon dioxide see p.11 - Cross out HCO - (aq) on the species list, and add H 2 CO (aq). Since this is an acid (not a base), it is not the next strongest base, but water is, the final reaction. Na(aq) CO 2- (aq) H 2 O(l) HCO - (aq) H 2 CO (aq) Cl - (aq) H O (aq) s SB A/SB A/SB A s SA H 2 O(l) OH - (aq) H 2 O(l) OH - (aq) - To choose indicators for these reactions, look at the bases reacting in the Erlenmeyer flask. The first base is CO 2- (aq), the next strongest base is HCO - (aq), so an indicator may be chosen for this first equivalence point. Since the next base is HCO - (aq) and the base just consumed is CO 2- (aq), we can choose an indicator ½ way between these two bases using p.11. Similarly, the other acid present is water, so a second indicator may be chosen between HCO - (aq) and H 2 O(l). Acid Conj. Base Ka H O (aq) H 2 O 1 H 2 CO HCO - 4.5x10-7 Choose an indicator for the second equivalence point here (around 10-4 ) HCO - CO 2-4.7x10-11 Choose an indicator for the first equivalence point here (around 10-9 ) Chemistry 0 Unit VI Notes page 7

8 - Using these guidelines, an appropriate indicator for the first equivalence point would be phenolphthalein (Ph - (aq)), and for the second could be methyl orange (Mo - (aq)). - The indicators are in the basic form since they are in a flask with a stronger base. Now, complete with indicator reactions, the reactions for this titration are: CO 2- (aq) H O (aq) HCO - (aq) H 2 O(l) Ph - (aq) H O (aq) HPh(aq) H 2 O(l) pink colourless HCO - (aq) H O (aq) H 2 CO (aq) H 2 O(l) Mo - (aq) H O (aq) HMo(aq) H 2 O(l) yellow red H 2 O(l) OH - (aq) H 2 O(l) OH - (aq) - There are two colour changes for this reaction, the first will occur after the first equivalence point (pink to colourless) and the second after the second equivalence point (yellow to red). - Since the two indicators are put in the Erlenmeyer at the beginning of the experiment, the colours and their changes will be added together. Initially, the colours will be yellow and pink (the initial colours of both indicators which add to produce a peach colour). The phenolphthalein will change first to colourless, so the two colours together are colourless and yellow (yellow overall), and finally the yellow will change to red, so red overall (red and colourless). The overall endpoint is red. phenolphthalein Pink Colourless Colourless methyl orange Yellow Yellow Red overall Peach Yellow Red Ex: What is the if methyl red is red, methyl orange is yellow and bromocresol green is green? If methyl red is red, <4.8 If methyl orange is yellow, > 4.4 If bromocresol green is green, >.8 but <5.4 So, the must be between 4.4 and 4.8 Ex: At a of 8., what is the colour of a mixture of orange IV, phenol red and thymolphthalein? orange IV is yellow phenol red is red thymolphthalein is colourless So, the overall colour is yellow red orange. Chemistry 0 Unit VI Notes page 8

9 Part III - Titration Calculations There are two calculations you have to be able to do: - Stoichiometry (calculating concentration) - at any point in the titration of a strong acid and strong base (mixture problem) Calculating concentration (stoichiometry) - This is just stoich with a couple of warnings! *** be careful about the equivalence point you are using! *** make sure you are using the molar mass of the compound you really want and the molar ratio for what you really want! - It is probably easiest to use net ionic equations (since that s what you ve been using all unit). To get the net ionic equation, follow the steps we ve used all along list all of the species as they appear and react the strongest base with the strongest acid. Ex: A 50 m solution of a mobasic substance required 2.7 m of 100 mol/ HCl(aq) to reach equivalence. Determine the concentration of the unknown base. Since it is a base, we can use A - (aq) to represent the base. A - (aq) H 2 O(l) Cl - (aq) H O (aq) SB A/B s SA Net ionic equation: A - (aq) H O (aq) HA(aq) H 2 O(l) Now that you have the equation, use dimensional analysis to get the unit you want. xmol 100molHCl 027 1molHO A ( aq) 1 1molHCl 0474mol A ( aq) 1molA 1molH O This is a step you ll probably want to leave out just be careful that it really is a 1:1 ratio (I wouldn t leave it out if I were you) ex: What is the concentration of the phosphoric acid if 45. m of 200 mol/ NaOH(aq) was required to reach the 2 nd equivalence point in the titration of 25.0 m of the acid. Since the question states that the H PO 4 (aq) was titrated to the 2 nd equivalence point, you must remember that two titration steps are completed! (Write two net ionic equations and then add them up). H PO 4 (aq) H 2 O(l) H 2 PO - 4 (aq) Na (aq) OH - (aq) titrant SA A/SB A/B s SB H PO 4 (aq) OH - (aq) H 2 PO 4 - (aq) H 2 O(l) H 2 PO 4 - (aq) OH - (aq) HPO 4 2- (aq) H 2 O(l) When the H PO 4(aq) is used up, it produces H 2PO 4 - (aq), which is added to the list of reactants, and is the 2 nd strongest acid, so is used in the 2 nd reaction H PO 4 (aq) 2 OH - (aq) HPO 4 2- (aq) 2 H 2 O(l) This is the overall reaction to the 2 nd equivalence point (two steps added together) and gives you a molar ratio of 2 base: 1 acid Chemistry 0 Unit VI Notes page 9

10 xmol 200molNaOH 1molOH 045 1molHPO4 HPO4( aq) 1molNaOH 1 2molOH 181mol HPO 4 ( aq) Ex: Using the following titration curve, determine the concentration of HCl(aq). The molar ratio from the 2 nd equivalence point The titration of 25.0 m of HCl(aq) with 100 mol/ NaOH(aq) - Use the information in the title and the equivalence point (the midpoint of the steepest part of the graph) to determine the values needed for the the calculation m of 100 mol/ NaOH(aq) added to 25.0 m HCl(aq) 14.7 Volume of NaOH added (m) H O (aq) Cl - (aq) H 2 O(l) Na (aq) OH - (aq) (titrant) SA s A/B s SB H O (aq) OH - (aq) 2 H 2 O(l) xmol HCl( aq) 100molNaOH 1molOH molH O 1molNaOH 1 1molOH 1molHCl 1molH O mol HCl( aq) Ex: Use the following titration curve to determine the concentration of Na 2 CO (aq). The titration of 10 m of Na 2CO (aq) with 100 mol/ HCl(aq) - When titrating polyprotic acids and bases, you can use either the first equivalence point volume and write a reaction for the first equivalence point, or use the volume for the second equivalence point and write the reaction for the second they will both give the same answer since the 2 nd volume is double the first, but the molar ratio is ½ Volume of HCl added (m) Chemistry 0 Unit VI Notes page 10

11 For the first equivalence point: Na(aq) CO 2- (aq) H 2 O(l) Cl - (aq) H O (aq) s SB A/B s SA CO 2- (aq) H O (aq) HCO - (aq) H 2 O(l) xmol Na CO ( aq) 2 100molHCl 1molHO 1molHCl molCO 1 1molH O 2 1molNa2CO 2 1molCO mol Na2CO ( aq) For the second equivalence point: Na(aq) CO 2- (aq) H 2 O(l) HCO - (aq) Cl - (aq) H O (aq) s SB A/B A/B s SA CO 2- (aq) H O (aq) HCO - (aq) H 2 O(l) HCO - (aq) H O (aq) H 2 CO (aq) H 2 O(l) CO 2- (aq) 2 H O (aq) H 2 CO (aq) 2 H 2 O(l) Double the volume to the second equivalence point xmol Na CO ( aq) 2 100molHCl 1molH O 1molHCl molCO 1 2molH O 2 1molNa2CO 2 1molCO mol Na2CO ( aq) Molar ratio changes to half that of the first equivalence point. - Sometimes the titration numbers will be provided in a table of lab data. Don t forget to omit any data point that is more than 20 m away from the other points (if they are close together). Ex: Determine the concentration of HCl(aq) given the following data in the titration of 25.0 m samples of acid with 15 mol/ NaOH(aq) Trial 1 Trial 2 Trial Trial 4 Final volume (m) Initial volume (m) Volume added (m) The volume that should be used in the calculation is m (the average of trials 2-4). Chemistry 0 Unit VI Notes page 11

12 Calculating during a titration of a strong acid/strong base Calculating the during a titration is just a mixture problem with the quantity of acid and base provided from the titration. The at every point in a titration can be calculated this way. Ex: In the titration of 20 m of 200 mol/ NaOH(aq) with 100 mol/ HCl(aq), determine the at: a) 0 m HCl(aq) added b) 20 m HCl(aq) added c) 40 m HCl(aq) added d) 60 m HCl(aq) added (To refresh your memory, to do a mixture problem, find the moles each of H O (aq) and OH - (aq) present, subtract them to get the excess moles, divide by the total volume and then calculate ) a) When no HCl(aq) has been added, only NaOH(aq) is present. [OH - (aq)] 200 mol/, so poh -log(200) poh b) When 20 m of HCl(aq) has been added: mol H O (aq) 100 mol/ x mol mol OH - (aq) 200 mol/ x mol Total volume 0400 Excess OH - /H O mol OH - So [OH - (aq)] mol/ mol/ poh c) When 40 m of HCl(aq) has been added: mol H O (aq) 100 mol/ x mol mol OH - (aq) 200 mol/ x mol Total volume 0600 Excess OH - /H O 00 mol So [OH - (aq)] [H O (aq)] 1.00x x10-7 mol/ poh d) When 60 m of HCl(aq) has been added: mol H O (aq) 100 mol/ x mol mol OH - (aq) 200 mol/ x mol Total volume 0800 Excess OH - /H O mol H O So [H O (aq)] mol/ mol/ Chemistry 0 Unit VI Notes page 12