Chemistry 12 UNIT 4 ACIDS AND BASES

Transcription

1 Chemistry 12 UNIT 4 ACIDS AND BASES PACKAGE #6 TITRATION -allows you to react measured amounts of a solution with a known volume of another solution until an equivalence point is reached. Recall from Indicators notes: equivalence point - proportions of reactants in the beaker are equivalent to proportions described in your stoichiometric equation. There are many types of titrations, as we will witness in Unit 4 and Unit 5. However in Unit 4, we will be dealing with Acid-Base Titrations -an acidic solution of known concentration is added to a basic solution of unknown concentration or vice versa. The solution of known concentration is called the standard solution. To prepare a standard solution, you can accurately determine the mass of a pure dry substance and dissolve it in water, to the required volume. ( a little bit of water in the beaker/cylinder, add enough dry solute to satisfy molarity requirements, stir until dissolved and then add water to make up to accurate volume for molarity requirements). It is difficult to obtain pure dry acids and bases since some are extremely hygroscopic (water absorbing) such as NaOH pellets. This means you have a hard time accurately determining the mass of the dry NaOH. So you can use a primary standard -highly pure and non-hygroscopic. -reacts rapidly with the solution to be standardized. -of known concentration. Once we have a standard solution or primary standard, we can perform a titration. This titration will determine the volume of solution required to react with an exactly known amount of primary standard. We use an indicator in the titration to signal when we have reached the endpoint.

2 We always need a standard solution of some kind to accurately measure how much acid or base is required to react with the unknown base or acid. If you do not have a standard solution, you may standardize a solution as follows: How to Standardize a Solution: For example, consider that we are going to do an acid-base titration (titrating an acid with a base) and we do not have a basic primary standard, even though we need one for our titration. We have a basic solution, but we do not accurately know it's concentration. We would choose an available acidic primary standard to accurately determine our base concentration. Then we can use our newly "standardized" basic solution for our titration into the unknown acid solution. See notes on indicators for set up of burette and flask during titrations. Also see in these notes the discussion of how indicators are used to signal the endpoint, and if calculations are done correctly, this should equal the equivalence point. This is where our neutralization calculations come in handy. We need to know how to write formula, complete ionic, and net ionic neutralization equations for the neutralizations that occur during a titration.

3 TITRATION CURVES A TITRATION CURVE is a plot of ph versus the amount (volume in ml) of acid or base added. (recall that "ph versus volume" means ph on y axis and volume on x axis). It displays graphically, the change in ph as the acid or base is added to a solution and indicates clearly how ph changes near the equivalence point. STRONG ACID / STRONG BASE This is a titration in which you are adding a strong base to a strong acid. see graph p. 442 Chem text. This curve shows the titration of 25.0 ml of M HCl with M NaOH.

4 We can accurately determine the ph of the curve at different stages of the titration. When no base had been added yet: HCl is a strong acid: HCl (aq) + H2O (l) à H3O + (aq) + Cl - (aq) So most of the [H3O + ] comes from the HCl Therefore [H3O + ] = M ph = When 12.5 ml of base had been added: HCl (aq) + NaOH (aq) à NaCl (aq) + H2O (l) initial moles of HCl: 25.0 ml x M = moles = [H3O + ] (0.025 L) moles of NaOH added: 12.5 ml x M = moles = [OH - ] ( L) (moles NaOH added = moles of HCl reacted since 1:1 ratio) amount HCl remaining: = moles (unreacted) Since HCl is the major source of H3O +, and there is now a total volume of ml = 37.5 ml, then [H3O + ]= moles / 37.5 ml = M (0.0375L) ph = When 25.0 ml of base had been added: initial moles of HCl: moles = [H3O + ] moles of NaOH added: 25.0 ml x M = moles = [OH - ] We have reached the equivalence point: there are equal amounts of acid and base! moles [H3O + ] = [OH - ] Now the flask contains only the products of the reaction: NaCl (aq) + H2O (l) Since H2O is the major supplier of H3O + and OH -, 2H2O H3O + + OH -

5 Keq = [H3O + ][OH - ] [H2O] 2 Keq = [H3O + ][OH - ] Kw = [H3O + ][OH - ] but moles [H3O + ] = [OH - ] Kw = [H3O + ] 2 [H3O + ] = Kw = 1.0 x = 1.0 x 10-7 M ph = 7.00 When excess base had been added: 50.0 ml of M NaOH: initial moles of HCl: moles = [H3O + ] = amount of NaOH used in the neutralization reaction. BUT the amount of NaOH added = 50.0 ml x M = moles Excess [OH - ] = moles moles 25 ml HCl + 50 ml NaOH = M poh = ph = Let's look at the curve again: -the corners are quite steep -the ph rises vertically around the equivalence point (small additions of base cause the ph to change quite drastically in this range) (the midpoint of this section is called the equivalence point) -we call the amount of base required to reach the equivalence Vb in our calculations CHOOSING AN INDICATOR: We want to choose an indicator that changes colour within this vertical section, preferably at the calculated equivalence point of ph = So, bromthymol blue or phenol red.

6 STRONG BASE / STRONG ACID This is a titration in which you are adding a strong acid to a strong base.

7 YOU MUST be able to WRITE FORMULA, COMPLETE IONIC AND NET IONIC NEUTRALIZATION EQUATIONS FOR TITRATIONS / NEUTRALIZATIONS. FORMULA: HCl (aq) + NaOH (aq) à NaCl (aq) + H2O (l) COMPLETE IONIC: H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq) à Na+ (aq) +Cl- (aq) + H2O (l) NET IONIC: H+ (aq) + OH- (aq) à H2O (l)

8 WEAK ACID / STRONG BASE see graph p. 443 Chem text. This curve shows the titration of 28.0 ml of M CH3COOH with M NaOH. Notice that the ph at the equivalence point is basic. (due to the strong base) The ph at the equivalence point will be different depending on the concentration and volumes of the acid and the base. (In this case, the ph will not work out to 7.00, because K will not necessarily be Kw or Kn). A common indicator for this category of titrations is phenolphthalein. It has a broad range, and it's colour change is VERY easy to detect. FORMULA: CH3COOH (aq) + NaOH (aq) à NaCH3COO (aq) + H2O (l) COMPLETE IONIC: H + (aq) + CH3COO - (aq) + Na + (aq) + OH - (aq) à Na (aq) +CH3COO - (aq) + H2O (l) NET IONIC: H + (aq) + OH - (aq) à H2O (l)

9 STRONG ACID / WEAK BASE see graph p. 443 Chem text. This curve shows the titration of 80.0 ml of M HCl with M NH3. (or sometimes written as NH4OH which is NH3 + H2O) Notice that the ph at the equivalence point is acidic. (due to the strong acid) For this particular titration, the ph at the equivalence point seems to be around 6.50 or more. A suitable indicator would be FORMULA: HCl (aq) + NH4OH (aq) à NH4Cl (aq) + H2O (l) COMPLETE IONIC: H + (aq) + Cl - (aq) + NH4 + (aq) +OH - (aq)à NH4 + (aq) + Cl - (aq) + H2O (l) NET IONIC: H + (aq) + OH - (aq) à H2O (l)

10 WEAK BASE / STRONG ACID

11 TITRATION CURVES A SUMMARY - COMPARISON STRONG / STRONG (acid) (base) -the ph initially starts out as a plateau or a very gradual positive slope -the corners are steep -the ph rises vertically around the equivalence point (the midpoint of the vertical section is the equiv. point) WEAK / STRONG (acid) (base) -there is an initial upswing in the ph at the start of the titration -the corners are gradual -the ph rises gradually at equiv. pt. (more of a sloped or diagonal line versus the straight line seen in strong / strong) -small additions of base cause the ph to change quite drastically around the range of the equivalence point -the ph at the equiv. point is 7 -the ph at the equiv. pt is >7 QUESTION: Why does the slope of the curve gradually decrease as we pass the equiv pt and get to the point where too much base had been added? I.e. Why doesn't the ph keep shooting upward as more and more base has been added?? ANSWER: Every acid and base has a certain ph / poh value. That is why on the exam they ask you to calculate the ph of M HNO2 or the poh of M NH3. Therefore, each base has a sort of a "limit" to how basic it can actually get, depending on its concentration, strength, and possibly even how much the total volume of base + acid is.

12 NEUTRALIZATION: Until now, we were satisfied to write: ACID + BASE à SALT + WATER In Chemistry 12, we tend to write the NET IONIC EQUATION for a neutralization: H3O + (aq) + OH - (aq) à 2H2O (aq) Keq = 1/Kw = 1/10-14 = K is very large therefore the reaction is essentially complete (100% to the right) In any neutralization the number of moles of H3O + = the number of moles of OH - (consider M1V1= M2V2 OR MaVa = MbVb) Sample Calculations: 1) Calculate the volume of 0.10 M sulphuric acid needed to neutralized 50.0 ml of 0.25 M sodium hydroxide solution. 2) 4.0 g of calcium hydroxide is dissolved in 50.0 ml of water. How many ml of 0.10 M hydrochloric acid is needed to neutralize this solution? 3) What volume of 0.34 M KOH will neutralize 65 ml of 0.25 M HCl? 4) Find the [HCN] if 75.0 ml is neutralized by 15 ml of 0.60 M LiOH. What indicator would be suitable? 5) If 30.0 ml of 0.16 M sulphuric acid is neutralized by 54.0 ml of KOH, find [KOH]. 6) 50.0 ml of M NaOH is mixed with 50.0 ml of M HCl. Calculate the ph of the resulting solution. 7) ml of a solution containing 6.08g of Sr(OH)2 is mixed with ml of a solution containing 8.09g of HBr. Calculate the poh. 8) What mass of Ca(OH)2 must be added to ml of a M HBr solution to result in a solution of ph = 2.750? Please note that some of these calculations will be tougher to complete until the Titration notes package has been completed.

13 NEUTRALIZATION Sample Calculations ANSWERS ml L ml 4. This stoichiometric point will have ph > 7 since we've combined equal amounts of weak acid with strong base. Bromothymol blue would NOT be appropriate for this reaction. Phenolphthalein could be used. 5. [KOH] = 0.17 M 6. ph is Resulting solution is neutral. Therefore poh = ph = g Ca(OH)2

14 NEUTRALIZATION Sample Calculations FULL SOLUTIONS: 1) (0.1M x 2)(Va)= (0.25N)(50.0 ml) = 62.5 ml (2 N ) 2) Assume no volume change 4.0 g / 74 g / mole Ca(OH)2 = moles moles / 50.0 ml = 1.08 M Nb = 2.16 N = 2.2 N (2x 1.08 M) Va = 1.08 L NOTE THAT THE VOLUME DID NOT MATTER (does not matter if dissolved in 50 ml, 100 ml, 1000 ml etc, it just matters how many moles were dissolved) IN THIS CASE, WE COULD USE: the number of moles of H3O + = the number of moles of OH - NaVa = M Va = / 0.1 N = 1.08 L 3) [H + ] = [OH - ] 65 (0.25M) (0.34M)y y + 65 y + 65 y = 47.8 ml 4) [HCN] = [OH - ] 75.0 ml [HCN] = [0.60M] 15 ml 90.0 ml 90.0 ml [HCN] = 0.12 M This stoichiometric point will have ph > 7 since we've combined equal amounts of weak acid with strong base. Bromothymol blue would NOT be appropriate for this reaction. Phenolphthalein could be used. 5) [H + ] = [OH - ] 30.0 ml(0.32m) 54.0 ml[koh] [KOH] = 0.17 M (note that H2SO4à 2H + + SO4 2- ) 0.16M 0.32M

15 6) TAKEN FROM HEBDEN P. 143 # 58 HCl: M1V1 = M2V2 (do the dilution calc first) (50.0)(0.200) = M2(100) M2= M NaOH: M1V1 = M2V2 (do the dilution calc first) (50.0)(0.150) = M2(100) M2= M In a neutralization: H + + OH - <====> H2O XS limiting Or consider it as follows: H + + OH - <====> H2O I R E The amount of excess or unreacted H + is M (This is equal to the [H + ] initial - [OH - ] initial From this [H + ] the ph is (It should be acidic since there was XS H + ) 7) TAKEN FROM HEBDEN P. 143 # 63 HBr: M1V1 = M2V2 (do the dilution calc first) M1 = 8.09g x (1 mole / 80.9g) 100ml (8.09g x (1 mole / 80.9g) (100.0 ml) = M2(350.0 ml) 100ml (.3500L) M2= M Sr(OH)2: M1V1 = M2V2 (do the dilution calc first) (6.08g x (1 mole / 121.6g) (250.0 ml) = M2(350.0 ml) 250.0ml (.3500L) M2= M THIS IS THE CONCENTRATION OF THE Sr(OH)2 HOWEVER THE [OH - ] IS M x 2 = 0.286M In a neutralization: H + + OH - <====> H2O [H + ] = [OH - ] Therefore neither ion is in excess Therefore the solution is neutral. ph = 7

16 8) TAKEN FROM HEBDEN P. 143 # 67 ph = [H + ] = = 1.78 x 10-3 M This is the excess [H + ] In a neutralization: H + + OH - <====> H2O XS 1.78 x 10-3 limiting actually x reacted Or consider it as follows: H + + OH - <====> H2O I 1.78 x 10-3 R E x NOTE FROM QUESTIONS 6 AND 7: [H + ] XS = [H + ] initial - [OH - ] initial [OH - ] XS = [OH - ] initial - [H + ] initial Therefore: [H + ] XS = [H + ] initial - [OH - ] initial 1.78 x 10-3 = x x = M = [OH - ] initial Ca(OH)2 <====> Ca +2 + OH using mole ratios, [Ca(OH)2] = M M Ca(OH)2 x 0.500L x 74.1 g/mole = g Ca(OH)2