Chapter 16 Cartoon from: Strong Acid/Base Neutralization

Transcription

1 Chapter 16 Applications of Aqueous Equilibria Chapter 16 Cartoon from: 1 Strong Acid/Base Neutralization When a strong acid and a strong base react, the products are always water and a salt. The formula of the salt depends on the anion of the acid and the cation of the base. The overall ph of the solution will be neutral because the salt does not effect the ph and water is neutral! H + (aq) + OH (aq) H 2 O (l) K = = C + - [H ][OH ] K Net Ionic Reaction Chapter 16 2 W 1

2 Strong Acid/Base Neutralization To write the products of a neutralization reaction, you MUST know how to put together an ionic compound (Review Section 2.10) Complete and Balance the following reactions: H 3 PO 4 (aq) + Ca(OH) 2 (aq) HNO 3 (aq) + NaOH(aq) HClO 4 (aq) + Mg(OH) 2 (aq) Chapter 16 3 Predicting Neutralization Reactions We can also identify the strong acid and base that react in a neutralization reaction to produce a given salt. For example, what strong acid and strong base react to give the salt, calcium sulfate? Strong Acid + Strong Base CaSO 4 (aq) + H 2 O(l) The calcium must be from a strong base, calcium hydroxide. The sulfate must be from a strong acid, sulfuric acid H 2 SO 4 (aq) + Ca(OH) 2 (s) CaSO 4 (aq) + 2 H 2 O(l) Chapter

3 Neutralization: Weak Acid / Strong Base When a weak acid and a strong base react, the products are also a water and a salt. Again, the formula of the salt depends on the anion of the acid and the cation of the base. However, in this case, the anion of salt will effect the ph of the solution. In other words, the ph is not neutral HCN (aq) + OH - (aq) H 2 O (l) + CN - (aq) [CN ] - Net Ionic Reaction a K = = C - [HCN][OH ] K W Chapter 16 5 K Neutralization: Strong Acid / Weak Base When a strong acid and a weak base react, the products are also a water and a salt. Again, the formula of the salt depends on the anion of the acid and the cation of the base. However, in this case, the cation of the salt will effect the ph of the solution. In other words, the ph is not neutral H 3 O + (aq) + NH 3 (aq) H 2 O (l) + NH 4+ (aq) [NH ] [H O ][NH + 4 K = = C + 3 Chapter ] K K b W Net Ionic Reaction 3

4 Neutralization: Weak Acid / Weak Base When a weak acid and a weak base react, the products are only a salt. The formula of the salt depends on the anion of the acid and the cation of the base. In this case, the both the cation and anion of the salt will effect the ph of the solution. In other words, the ph is not neutral HCN (aq) + NH 3 (aq) CN - (aq) + NH 4+ (aq) K C - + [CN ][NH ] K K 4 b a = = [HCN][NH ] K Chapter 16 3 W 7 Predicting Neutralization Reactions Write a balanced net ionic equation for the neutralization of the following acids and bases. Indicate whether the ph after neutralization is acidic, basic or neutral. HNO 2 and KOH HF and NH 3 HBr and NH 3 Chapter

5 A buffer is a solution that resists changes in ph when an acid or a base is added. A buffer always contains a combination of an acid-base conjugate pair. For example, a buffer can be made up from equal concentrations of a weak acid and a salt of the conjugate base of that acid. May also contain a weak base and a salt with the conjugate acid. Buffers Chapter 16 9 Identifying Buffer Solutions When trying to determine whether a solution is a buffer, you need to determine: 1) whether a conjugate acid-base pair is present; and 2) whether that pair consists of weak acids/bases. State whether the following pairs will be a buffer: NaCl and Na 2 CO 3 NaCl and HCl NaF and HF CH 3 COOH and NaCH 3 COO Chapter

6 Buffer Capacity Buffer Capacity is a measure of the amount of acid or base that a buffer system can neutralize while maintaining its own ph The buffer capacity depends on the number of moles of the conjugate pair. As the concentration of the buffer species increases, the buffer capacity increases. Chapter Calculating the ph of a Buffer We cannot use the ph formula to determine the ph of a buffer because there is an acid AND a base present. We use the Henderson-Hasslebach Equation for this calculation. ph = pk a + log [Base] [Acid] When the [Base]/[Acid] = 1, then ph = pk a Conjugate Pair Chapter

7 Calculating the ph of a Buffer Calculate the ph of a buffer system containing 1.0 M CH 3 COOH and 1.0 M CH 3 COONa. What is the ph of the system after the addition of 0.10 mole of gaseous HCl to 1.0 L of solution? Calculate the ph of 0.30 M NH 3 /0.36 NH 4 Cl buffer system. What is the ph after the addition of 20.0 ml of M NaOH to 80.0 ml of the buffer solution? Chapter To prepare a buffer: Buffer Preparation 1) Choose weak acid with pk a close to required ph. 2) Substitute into Henderson Hasselbalch equation. 3) Solve for the ratio of [conjugate base]/[acid]. This will give the mole ratio of conjugate base to acid. You can use this ratio and the volume of your solution to calculate the amount of each species you will need to prepare the buffer. How would you prepare 1.00 L of carbonate buffer at a ph of 10.10? You are provided with carbonic acid (H 2 CO 3 ), sodium hydrogen carbonate (NaHCO 3 ), and sodium carbonate (Na 2 CO 3 ). Chapter

8 Titrations Chapter Acid-Base Titrations An acid-base titration is used to analyze an acid solution using a solution of a base (with a known concentration) or vice versa. A measured volume of base of known concentration is added to an acid solution of unknown concentration. Titrations involving only strong acids and bases are straightforward Titrations involving weak acids and/or bases are complicated by the formation of a salt that effects ph When a stoichiometric amount of acid and base have reacted, you have reached the endpoint or equivalence point. We use an indicator or a ph meter to show us when the solution reaches the endpoint. Chapter

9 Strong Acid/Base Titration There are four stages of a titration: A) Before the addition of any titrant: Here, the ph is dependent only on the amount of acid or base being titrated B) Before the equivalence point: Here, you must determine how much of the acid or base being titrated has been neutralized (by subtracting added from initial), then calculating ph. C) At the equivalence point: For a strong acid/base titration, the ph = 7.0 at the equivalence point. D) Beyond the equivalence point: Now, the ph is dependent solely on the excess titrant that has been added beyond the Chapter A D B ph at equivalence point will always be 7.0 C Strong Acid/Base Titration What is the ph of a 25.0 ml 0.10 M HCl sample after each of the following additions: No addition of 0.10 M NaOH ml (total) of 0.10 M NaOH ml (total) of 0.10 M NaOH ml (total) of 0.10 M NaOH. Chapter

10 Titration of a Weak Acid with a Strong Base There are four stages of a titration: A) Before the addition of any base: Here, the ph is dependent only on the amount of weak acid. B) Before the equivalence point: Here, both the weak acid and its conjugate will be present. You must determine how much of the acid has been converted to its conjugate, then use the H-H equation to calculate ph. C) At the equivalence point: Here, the acid has been completely converted to its conjugate base. The ph is dependent on the amount of CB present. D) Beyond the equivalence point: Now, there is excess strong base in solution. The ph is dependent solely on this excess base. The CB has a negligible effect and is ignored. Chapter A C B D ph at equivalence point will always be > 7.0 Weak Acid / Strong Base Titration The ph of a 25.0 ml 0.10 M CH 3 COOH sample can be determined after the addition of: No addition of 0.10 M NaOH ml (total) of 0.10 M NaOH ml (total) of 0.10 M NaOH ml (total) of 0.10 M NaOH. Chapter

11 Titration of a Strong Acid with a Weak Base There are four stages of a titration: A) Before the addition of any acid: Here, the ph is dependent only on the amount of weak base. B) Before the equivalence point: Here, both the weak base and its conjugate will be present. You must determine how much of the base has been converted to its conjugate, then use the H-H equation to calculate ph. C) At the equivalence point: Here, the base has been completely converted to its conjugate acid. The ph is dependent on the amount of CA present. D) Beyond the equivalence point: Now, there is excess strong acid in solution. The ph is dependent solely on this excess acid. The CA has a negligible effect and is ignored. Chapter A C B D ph at equivalence point will always be < 7.0 Strong Acid / Weak Base Titration The ph of a 25.0 ml 0.10 M NH 3 sample can be determined after the addition of: No addition of 0.10 M HCl ml (total) of 0.10 M HCl ml (total) of 0.10 M HCl ml (total) of 0.10 M HCl. Chapter

12 Polyprotic Acids Chapter Solubility Equilibria Many biological processed involve the dissolution or precipitation of slightly soluble ionic compounds. M m X x (s) m M x+ (aq) + x X m- (aq) Rather than writing a K c expression, we write a solubility product constant (K sp ) for saturated solutions. K SP = [M x+ ] m [X m ] x This expression can only be written for the insoluble salts (which are slightly soluble) WHY?! Chapter

13 Solubility Product Equilibria We write the K sp the same way we do the K c Remember, we don t include pure solids or liquids in equilibrium constant expressions. If we add Ag 2 SO 4 to water, some of the slightly soluble Ag 2 SO 4 dissolves: Ag 2 SO 4 (s) 2 Ag + (aq) + SO 2-4 (aq) We can write the K sp expression for the reaction: K sp = [Ag + ] 2 [SO 2-4 ] Write the K SP expression for the following: AgCl (s) Ca 3 (PO 4 ) 2 (aq) PbI 2 (s) Cr(OH) 3 (s) Chapter Calculation of K sp For the following equilibrium, the concentrations are: [Mg 2+ ] = M [OH - ] = M. What is K sp? K sp = [Mg 2+ ][OH - ] 2 = ( )( ) 2 K sp = Chapter

14 K sp Values How do you think the K SP relates to a compound s solubility? Chapter 16 Many other K SP values in Table C.4 in the Appendix 27 Calculation of Solubility Once a K SP value is know for a compound, you can use it to calculate the solubility of the compound. These solubilities are approximate values (as there can be side reactions). Initial Change Equilibrium Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2- (aq) Excess - X Excess + 2X K sp = [Ag + ] 2 [CrO 2-4 ] = (2X) 2 (X) X = 1.4 x 10-4 M 0 + X X Chapter X 14

15 Ion Solubility Product (Q SP ) The Ion Solubility Product (Q SP ) is the solubility equivalent of the reaction quotient. It can be used to determine whether a precipitate will form. Q < K sp Q = K sp Q > K sp Unsaturated Saturated Supersaturated; ppt forms. Chapter Solubility Product Equilibria Exactly 200 ml of M BaCl 2 are added to exactly 600 ml of M K 2 SO 4. Will a precipitate form? If 2.00 ml of M NaOH are added to 1.00 L of M CaCl 2, will precipitation occur? Chapter

16 Fractional Precipitation Fractional precipitation is a method of removing one ion type while leaving others in solution. Ions are added that will form an insoluble product with one ion and a soluble one with others. When both products are insoluble, their relative K sp values can be used for separation. Chapter Fractional Precipitation Chapter

17 Fractional Precipitation Silver nitrate is slowly added to a solution that is M in Cl ions & M in Br ions. Calculate the concentration of Ag + ions (in mol/l) required to initiate (a) the pptn of AgBr, and (b) the pptn of AgCl. K sp s of AgCl and Ag 3 PO 4 are 1.6 x and 1.8 x 10 18, respectively. If Ag + is added to 1.00 L of a solution containing 0.10 mol Cl and 0.10 mol PO 3 4, calculate the concentration of Ag + ions required to initiate (a) the pptn of AgCl, and (b) the pptn of Ag 3 PO 4. Chapter Common Ion Effect and Solubility As the solubility product (K sp ) is an equilibrium constant, precipitation will occur only when the ion solubility product (Q) exceeds the K sp for a compound. This means that LeChatelier s principle is in effect. Increasing the concentration of one of the ions by adding a second solution puts stress on the equilibrium. PbCl 2 (s) Pb 2+ (aq) + 2 Cl - (aq) If NaCl is added to saturated PbCl 2, what will happen? What about adding NaNO 3? Chapter

18 Common Ion Effect and Solubility Calculate the solubility of silver chloride in a 6.5 x 10 3 M silver chloride solution. Calculate the solubility of AgBr in: (a) pure water (b) M NaBr Chapter ph and Solubility An ionic compound that contains a basic anion becomes more soluble as the ph decreases. CaCO 3 (s) Ca 2+ (aq) + CO 3 2- (aq) Chapter

19 Complex Ions and Solubility A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. Most metal cations in complex ions are transition metals because they have more than one oxidation state. The formation constant (K f ) is the equilibrium constant for the complex ion formation. The formation of a complex ion is a step-wise process and each step has its own equilibrium constant value: Ag + (aq) + NH 3 (aq) Ag(NH 3 ) + (aq) K 1 = 2.1 x 10 3 Ag(NH 3 ) + (aq) + NH 3 (aq) Ag(NH 3 ) 2 + (aq) K 2 = 8.1 x 10 3 Ag + (aq) + 2 NH 3 (aq) Ag(NH 3 ) 2 + (aq) K f = 1.7 x 10 7 What does this large value tell us about the formation of the complex ion? Chapter Complex Ions and Solubility The net reaction for the dissolution of AgCl in ammonia is an addition of the two net ionic equations: AgCl (s) Ag + (aq) + Cl - (aq) K SP = 1.8 x Ag + (aq) + 2 NH 3 (aq) Ag(NH 3 ) 2 + (aq) K f = 1.7 x 10 7 AgCl(s) + 2 NH 3 (aq) Ag(NH 3 ) 2 + (aq) + Cl - (aq) The value of K is much higher than the K SP value for AgCl, indicating that the solubility of the compound is greatly increased in ammonia versus water K = K SP x K f = 3.1 x 10-3 Chapter

20 Complex Ions and Solubility Calculate the molar solubility of AgBr in a 1.0 M NH 3 solution. The K SP of AgBr is 5.4 x The K f for the complex ion is 1.7 x A 0.20 mole quantity of CuSO 4 is added to a liter of 1.20 M NH 3 solution. What is the concentration of Cu 2+ ions at equilibrium? Several K f values in Table C.6 in the Appendix Chapter