Chapter 15 Equilibria in Acid-Base Solutions. AP Chemistry

Transcription

1 Chapter 15 Equilibria in Acid-Base Solutions AP Chemistry

2 Buffers Any solution containing appreciable amounts of both a weak acid and its conjugate base and: is highly resistant to change in ph brought about by addition of strong acid or strong base. has a ph close to the pka of the weak acid.

3 Characteristics of a buffer: It contains two species One that can react with H + One that can react with OH - [HB] eq [HB] o and [B - ] eq [B - ] o The ph of a buffer is independent of volume. [H + ] = K a [HB] = K a (moles HB) [B - ] (moles B - ) - Henderson-Hasselbach equation ph = pk a + log [B - ] [HB]

4 Buffer Systems

5 What happens when a buffer system in the body is overwhelmed?

6 Determination of H + (ph) Example #1: Calculate the ph of a solution with a volume of L that contains 2.70 g of HCN and 2.45 g of NaCN. K a for HCN is 5.8 x Solution: There are really three equilibrium systems involved in every buffer, for this buffer they are:

7 HCN (aq) H + (aq) + CN - (aq) H 2 O H + (aq) + OH - (aq) CN - (aq) + H 2 O HCN (aq) + OH - (aq) Always choose the equilibrium system for the ionization of the weak acid. That gives [H + ] directly, and then ph. The ionization of water can be ignored since the concentrations of H + and OH - are negligible compared to the same concentrations in the buffer system.

8 Write the ionization constant expression for HCN in terms of [H+]: [H+] = Ka [HCN] [CN-] = (5.8 x 10-10)(mole HCN) (mole CN-) moles HCN = 2.70 g (1 mole) = (27.03 g) moles CN- = moles NaCN = 2.45g(1mole) = (49.0 g)

9 Now, Substitute in: [H + ] = (5.8 x )( mole HCN) ( mole CN - ) [H + ] = 1.2 x 10-9 M Then ph = - log [H + ] so ph = 8.94 for the buffer. Example 15L.1, 2

10 Example #2: A H 2 PO 4- /HPO 4 2- buffer with a ph of 7.00 needs to be prepared. Taking the K a of H 2 PO 4- (H 2 PO 4- H + (aq)+hpo 4 2- (aq) ) to be 6.2 x 10-8, how many grams of NaH 2 PO 4 and Na 2 HPO 4, should be added to water to make this buffer? Solution: [H+] = = 10-7 M Use the ionization constant expression of the weak acid to calculate the weak acid-conjugate base ratio. K a = [H + ][HPO 4 2- ] [H 2 PO 4- ] Rearranging and substituting: [H 2 PO 4- ] = [H + ] = 1 x 10-7 = 1.6 [HPO 4 2- ] K a 6.2 x 10-8

11 This means that for every mole of weak base (HPO 4 2- ), we need 1.6 moles of weak acid (H 2 PO 4- ). Now calculate the mass of (Na 2 HPO 4 ). Since one mole of HPO 4 2- is needed, we also need one mole of Na 2 HPO 4. Mass Na 2 HPO 4 =1.0 mole(142 g)= 140 g (1 mole) Calculate the mass of Na 2 HPO 4 required. 1.6 moles of weak acid H 2 PO 4- is needed, so 1.6 moles of Na 2 HPO 4 is also needed. The mass is: Mass Na 2 HPO 4 = 1.6 mole(120 g) = 190 g (1 mole)

12 Therefore, to get a buffer of ph 7.0, dissolve 140 grams of Na 2 HPO 4. and 190 grams of NaH 2 PO 4. It doesn t matter how much water is used because the ph of a buffer is not affected by the volume of the solution. Example 15L.4

13 Adding H + or OH - to a buffer 1. Adding H + a. Recall that a buffer has one species that can react with H + ions. This species is the conjugate base B -. When H + is added to a buffer, the following reaction occurs. H + (aq) + B - (aq) HB (aq)

14 b. Addition of H + does two things to the buffer: 1 It makes more acid. The concentration of the weak acid in the buffer system after the addition of H + is Total moles HB= (moles HB) o + moles H + added 2 It uses up conjugate base. The reaction above shows that for every mole of H + added, an equal amount of weak base reacts. Therefore, the concentration of weak base (or conjugate base) after H + addition is Total mole B - = (moles B - ) o mole H + added

15 2. Adding OH - a. A buffer also has a species that can react with OH - ions. This species is the weak acid. The reaction is HB (aq) + OH - (aq) B - (aq) + H 2 O

16 b. Addition of OH - ion does two things to a buffer: 1 It makes more base. Each mole of OH - added makes an equivalent amount of weak base B -. Total moles B - =(moles B - ) o + moles OH - added 2 It uses up weak acid. Each mole of OH - added uses up an equivalent amount of weak acid. Total moles HB=(moles HB) o moles OH - added

17 Buffer limitations The buffer can absorb the large ph changes that adding strong acid or strong base imposes on a system as long as it has enough of the species to react with H + or OH -. A buffer that has mole HB and mole B -, for example, can only take up to mole H +, or OH -. After that, if say mole H + is added, then mole of the H + is unreacted and the ph of the resulting solution is reflected in the concentration of the H + in the system that now has mole H +.

18 Example #3 A buffer is prepared by mixing 3.00 g of benzoic acid (HC 7 H 5 O 2 ) and 4.50 g of sodium benzoate (NaC 7 H 5 O 2 ) in 2.00 L of water. The K a for benzoic acid is 6.6 x a. Calculate the ph of the buffer. Solution: Use the equilibrium system of the weak acid ionization. HC 7 H 5 O 2 H + (aq) + C 7 H 5 O 2 - (aq) The equation to determine H + is [H + ] = K a (mole HC 7 H 5 O 2 ) (mole C 7 H 5 O 2- )

19 Now find the number of moles of weak acid and conjugate base. Mole HC 7 H 5 O 2 = 3.00 g(1 mole HC 7 H 5 O 2 ) ( 122 g HC 7 H 5 O 2 ) Mole NaC 7 H 5 O 2 = 4.50 g(mole NaC 7 H 5 O 2 ) HC 7 H 5 O 2 = mole (144 g NaC 7 H 5 O 2 ) NaC 7 H 5 O 2 = mole Calculating [H + ] = 6.6 x 10-5 (0.0246) (0.0313) [H + ] = 5.2 x 10-5 M ph = 4.28

20 b. Calculate the ph of the buffer when 10.0 ml of M HCl is added. Solution: First calculate the number of moles of H + added L(0.100 M)(1 mole H + ) = 10-3 mole H + (1 mole HCl) Since H + is added, we increase the number of moles of weak acid by mol. mole HC 7 H 5 O 2 = =

21 The addition of H + decreases the number of moles of conjugate base by mole. Mole C 7 H 5 O 2- = = Now calculate the ph of the buffer after the addition of HCl. [H + ] = 6.6 x 10-5 (0.0256) = 5.6 x 10-5 (0.0303) The ph of the buffer after the addition of HCl is Recall that before HCl was added, the ph was 4.28.

22 c. Calculate the ph of the buffer when g of NaOH is added to the buffer. Solution: Calculate the moles of NaOH added, which is equal to the moles of OH - added g NaOH(1 mole NaOH)(1mole OH - ) = (40.0 g NaOH)(1 mole NaOH) = mole OH - Remember that the addition of OH - increases the number of moles of conjugate base and decreases the number of moles of weak acid: [H + ]= 6.6x10-5 ( )= 1.8 x 10-5 M ( ) The ph of the buffer after the addition of NaOH is 4.74.

23 d. Calculate the ph of the buffer when ml of M HCl are added. Solution: Calculate the number of moles of strong acid (also equal to the number of moles of H + ) added L(2.000 mole HCl)(1 mole H + ) = (1 L )(1 mole HCl) = mole H +

24 There is not enough weak base C 7 H 5 O 2- to react with H +. Since there is only mole of weak base, take = mole of unreacted H +. The buffer has been destroyed. The concentration of H + in the resulting solution: [H + ] = = M (2.000 L L) The ph of the solution is Note the drop from the original 4.28 ph. Example 15L.3, 5

25 Acid-Base Indicators Nature of an indicator 1. What does it do? It is useful in determining the equivalence point of an acid-base titration. If an indicator is chosen properly, the point at which it changes color (its end point) coincides with the equivalence point. 2. What is it derived from? From a weak acid HIn: HIn (aq) H + (aq) + In - (aq) K a = [H + ][In - ] [HIn] Where the weak acid HIn and its conjugate base In - have different colors.

26 3. Color a. When [HIn]/[In - ] > 10: 1. The principal species is _HIn_. 2. The color observed is that of _acid_. b. When HIn]/[In - ] < 0.10: 1. The principal species is _In - _. 2. The color observed is that of _base_. c. When [HIn] [In - ]: the color observed is that of _an intermediate_ d. Color is dependent on 1. ph of the solution because ph pk a 2. K a of the indicator because [H + ] K a

27 Acid-Base Titrations: Strong acid strong base titration This is a neutralization reaction. Example: HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O Where the net ionic equation is: H + (aq) + OH - (aq) H 2 O The titration curve looks like: The equivalence point has a ph = 7.00

28 Species present in solution at the equivalence point: a salt and water Types of indicators that can be used for this titration: methyl red (5), bromothymol blue (7), phenolphthaliein (9) Calculating the ph of the solution at different points in the titration In these titrations concentrate on three points 1. Starting point when no titrant has been added. 2. Half-way to the equivalence point 3. At the equivalence point

29 Example #4 If ml of M HClO 4 is titrated with M KOH, find the ph of the solution at the following points. a. When zero titrant has been added. Solution: The titrant is the species that is being added. It is the reagent whose volume is not initially given, in this case KOH. Before KOH is added, we have ml of a strong acid. (0.200 M HClO 4 ). Recall from Chapter 14 that the [H + ] of a strong acid is simply its molarity. [H + ] = M and ph = 0.699

30 b. Half-way to the equivalence point Solution: In this case, it is a good idea to first treat acid and base separately. Determine the volume, molarity and number of moles of the titrant needed to react with half the number of moles of the first. H + (from HClO 4 ): volume = L Molarity = M OH - (from KOH) : Volume = Moles = (0.200) = Molarity = M Moles = To determine how many moles of OH - have to be added to reach half-neutralization, we take half the moles of H +.

31 In this case: Mole OH - = ( )/2 = We can then determine the volume of KOH added. Volume = moles = L moles/l Now put the acid and base together. The reaction is H + (aq) + OH - (aq) H 2 O Reacting moles of H + with moles of OH - means that all the OH - gets used up and the number of moles of H + left over is = moles The total volume of the mixture is L L = L

32 [H + ] in the solution after KOH has been added half-way to the equivalence point is [H + ] = moles = M L The ph of the solution is

33 c. At the equivalence point. Solution: No calculations are necessary. The ph at the equivalence point for a strong acid strong base titration is always 7.

34 Strong Acid Weak Base titration Example: H 3 O + (aq) + NH 3 (aq) NH 4 + (aq) + H 2 O Draw the Titration Curve: (pg 427)

35 The ph at the equivalence point is:<7 ( 5) The species that are present in solution at the equivalence point: The conjugate acid and water. What are the types of indicators that can be used for this titration? Methyl Red In general, for the titration of a weak base with a strong acid, the indicator should change color on the acid side of ph 7.

36 Some Acid-Base Indicators

37 How to Calculate the ph: 1. When no titrant has been added: a. The titrant is almost always the strong acid. Determining the ph before addition of the strong acid is simply a matter of determining the ph of a solution of weak base. (This was done in Chapter 14).

38 b. At half-neutralization: Half of the weak base (B - ) has been converted to its conjugate acid (HB) by the reaction: B - (aq) + H + (aq) HB (aq) [B - ] = [HB] and substituting into the equilibrium expression: K b = [OH - ][HB] [B - ] K b = [OH - ] ph can easily be calculated from this.

39 c. At neutralization: All the weak base (B - ) has been converted to HB. The method used here is similar to the calculation of the ph of a weak acid discussed in chapter 14. Remember!! The molarity of the weak acid formed is determined by taking the number of moles of weak base (which is equal to the number of moles of conjugate acid formed) and dividing by the total volume of the solution (volume of weak base + volume of strong acid titrant).

40 Example # 5: Consider the titration of sodium cyanide, NaCN, with HCl. Twenty-five ml of M NaCN are titrated with M HCl. a. Calculate the ph when no titrant has been added. Solution: Before titration, the weak base CN - ionizes in water according to the equation CN - (aq) + H 2 O OH - (aq) + HCN (aq) The equilibrium constant expression is K b = [OH - ][HCN] [CN - ] Use a table again, it is the easiest way.

41 Substitute into the K b expression: K b = x 2 CN - H 2 O OH - HCN (0.200 x) [ ] o Look up K b for the reaction in your text [ ] -x +x +x and making the assumption that [ ] eq x x X x<<0.200, Then 1.7 x 10-5 = x [OH - ] = x = M Check assumption: x100 = 0.90% Well under 5.5%! Assumption is ok. From [OH - ]: poh = 2.73 and ph = 11.27

42 b. At half-neutralization: Solution: K b = [OH - ]= 1.7 x 10-5 poh = 4.77 and ph = 9.23 c. At the equivalence point: Solution: The volume of HCl required to reach the equivalence point is needed. Remember that the same number of moles of acid must be added to the moles of weak base in the solution to reach the equivalence point. Mole H + = mole CN - = L(0.200 M) = moles Since [H + ] = [HCl], then the volume of HCl can be calculated.

43 V HCl = mole HCl ( 1 L HCl ) = L (0.250 mole HCl) Find the molarity of the weak acid (HCN) formed. Recall that the number of moles of HCN formed is equal to both the number of moles of weak base (CN - ) and the number of moles of strong acid (H + ). The volume of the resulting solution is the sum of the volumes of weak base and strong acid. V HCN = L (from CN - ) l (from HCl) = L Now calculate molarity: M = moles HCN = mole = M V HCl L Knowing the molarity of the weak acid, construct a table and proceed to calculate the ph of the weak acid.

44 Substitute the values at equilibrium into the equilibrium expression: K a = [H + ][CN - ] [HCN] HCN H + CN - [ ] o [ ] -x +x +x [ ] eq x x x 5.8 x 10-5 = x(x) Assuming that x<<0.111 then (0.111-x) 5.8 x = x 2. Solving for x: x= 8.0 x 10-6 = [H + ] Checking assumption: 8.0 x 10-6 (100) = % << 5% so assumption is valid. The ph = -log(8.0 x 10-6 ) = 5.10

45 Weak acid-strong base titration: Example: HC 2 H 3 O 2 (aq) + OH - (aq) C 2 H 3 O - 2 (aq) + H 2 O Draw the titration curve (page 425).

46 The ph at the equivalence point is _>7 ( 9)_. The species present in solution at the equivalence point: Weak base The types of indicators that can be used for this titration are: Phenolphthalein (end point ph=9) For this type of titration choose an indicator that changes color above ph = 7!

47

48 Calculating the ph of the solution at different points in the titration: 1. When no titrant has been added: Determine the ph of the weak acid. (review chapter 14) 2. At half-neutralization: Use the same argument made in the strong acidweak base titration, that K a =[H + ] at halfneutralization. 3. At the equivalence point: A weak base is formed. [OH - ] is calculated using the ionization constant equation for K b. Remember the total volume of the solution (after an equivalent amount of titrant has been added) in order to calculate the molarity of the weak base.

49 Pollution can cause the environmental buffering systems to break down!

50 Equilibria in Acid-Base Solutions is used in everyday life! Without it we would not exist! Study and know these relationships! End of Chapter 15 STAY CAUGHT UP!