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Chapter 15 Applications of Aqueous Equilibria

Chapter 15 Applications of Aqueous Equilibria. Some of what we do in this chapter will be the same as what we did in Chapter 14. Let’s start with common ions. What do the weak acid HF and the salt NaF have in common?. The common ion effect.

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Chapter 15 Applications of Aqueous Equilibria

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  1. Chapter 15 Applications of Aqueous Equilibria • Some of what we do in this chapter will be the same as what we did in Chapter 14. • Let’s start with common ions. What do the weak acid HF and the salt NaF have in common?

  2. The common ion effect • When the salt with the anion of a weak acid is added to that acid: • it reverses the dissociation of the acid • Lowers the % dissociation of the acid • This same principle applies to salts with the cation of a weak base as well. • Calculations are the same as last chapter

  3. 15.2 Buffered solutions • When an acid-base solution contains a common ion, it’s called a buffered solution. • A buffered solution is one that resists a change in pH with the addition of hydroxide ions or protons. • Often buffered solutions contain a weak acid and it’s salt (HF and NaF) OR a weak base and it’s salt (NH3 and NH4Cl). • We can make a buffer of any pH by varying the concentrations of these solutions.

  4. Finding the pH of a buffered solution • Calculate the pH of a 1L buffered solution of 0.20M acetic acid solution and 0.30M sodium acetate. Ka=1.8x10-5

  5. pH changes in a buffered solution • We will use the solution from the last problem, but add 0.01M NaOH. Compare the pH change that occurs with the addition of the NaOH solid. • There are 2 major step to proceed with these types of problems!! • 1. The stoichiometry • 2. The equilibrium

  6. The stoichiometry1L of 0.20M acetic acid solution and 0.30M sodium acetate (Ka=1.8x10-5 ) [H+]=1.2x10-5 Buffered with .01M NaOH • Use moles not molarity • We will assume that all of the NaOH will be consumed

  7. Now for the equilibrium - ICE • Since there is 1L of solution

  8. pH change = 4.96 – 4.92 (from previous problem) Change in pH = .04 with the addition of a buffer.

  9. Think about what would have happened…. • If base was added to just water • That means the pH of 0.01M NaOH in water is 12 • Minus the pH of water (7) means a change in pH of 5. • Compare adding NaOH to a buffered solution vs. an unbuffered solution….5 vs 0.04…

  10. Remember • Buffered solutions are simply solutions of weak acids and weak bases containing a common ion • When a strong acid is added to a buffered soltuion…You know how to do these problems. Just remember the two steps… • Stoichiometry first • ICE (equilibrium) second

  11. How does the buffer work? • Solve the equilibrium expression for [H+] • This means the [H+] depends on the ratio of [HA]/[A-]…if you take the –log of both sides…(I won’t bore you with the math)… you get…

  12. It has a name! • It’s called the Henderson-Hasselback equation! • It also works for an acid and its salt, like HNO2 and NaNO2 • Or a base and its salt, like NH3 and NH4Cl, but you must remember to convert Ka to Kb in the equation

  13. An assumption • One assumption of this formula is that the initial concentrations and the equilibrium concentrations are equivalent. (<5% validity) • It is a pretty safe assumption since the initial concentrations of HA and A- are relatively large in a buffered system.

  14. Let’s try • What is the pH of a solution containing 0.30M HCOOH and 0.52M KCOOH (formic acid and potassium formate) Ka=1.8x10-4

  15. Given base, salt, and Kb • Calculate the pH of 0.25M NH3 and 0.40M NH4Cl. (Kb = 1.8 x 10-5) • Don’t forget to find Ka FIRST!! And the ratio is base over acid..

  16. 15.3 Buffer Capacity • The pH of a buffered solution is determined by the ratio of [A-]/[HA]. • If the ratio doesn’t change much…then the pH won’t change much either • The more concentrated these two are, the more H+ and OH- the solution will be able to absorb. • Larger concentrations = bigger buffer capacity

  17. Try this problem. • Calculate the change in pH that occurs when 0.040 mol of HCl(g) is added to 1.0 L of each of the following: Ka= 1.8x10-5 5.00 M HAc and 5.00 M NaAc 0.050 M HAc and 0.050 M NaAc • Calculate initial pH for each solution (henderson-hasselbalch)

  18. Now find the change after adding 0.04mol of HCl to 1.0L of solution • There’s virtually NO change. Now look at solution B. We already know the initial pH.

  19. Solution B is 1L of 0.050 M HAc and 0.050 M NaAc • Compared to 4.74 before the acid was added. Solution A contains much larger quantities of buffering components and has a larger buffering capacity than solution B

  20. Larger concentrations = bigger buffer capacity • Here’s why

  21. Buffer Capacity • The best buffers have a ratio [A-]/[HA] = 1 • This is the most resistant to change and true only when [A-] = [HA] • Makes pH = pKa (since the log of 1=0)

  22. 10.5 Titrations and pH Curves • Titration is commonly adding a solution of known concentration until the substance being tested is consumed. • This is often noted by a color change and is called the equivalence point. • This is often observed in a graph of pH vsmL of titrant and is called a titration curve.

  23. Strong acid with a strong base titration • Net ionic equation H+ + OH-H2O • To know the amount of H+ at any point in the titration, we must determine the amount of H+ remaining and divide by the total volume of the solution. • Let’s first consider a new unit for molarity that is smaller as most titrations usually deal with mL Millimole (mmol) = 1/1000 mol Molarity = mmol/mL = mol/L

  24. The pH Curve for the Titration of 50.0 mL of 0.200 M HNO3 with 0.100 MNaOH Where all the H+ ions originally present, have reacted with all the OH- ions added

  25. Things to note • You need to do the stoichiometry for each step • mL x Molarity = mmol • There is no equilibrium. Both the acid and base dissociate completely • Use [H+] or [OH-] to figure pH or pOH • The equivalence point is when you have an equal number of moles of H+ and OH- • In other words enough H+ is present to react exactly with the OH- • Before the equivalence point H+ is in excess • After the equivalence point OH- is in excess

  26. The pH Curve for the Titration of 100.0 mL of 0.50 MNaOH with 1.0 M HCI

  27. Strong base titrated with strong acid. • Very similar to strong acid with a strong base except before the equivalence point the OH- is in excess and H+ is in excess after the equivalence point.

  28. The pH Curve for the Titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 MNaOHweak acid – strong base At equivalence point (pH > 7): A- is basic

  29. Titrating a weak acid with a strong base • Again, there is no equilibrium • Do the stoichiometry – the reaction of OH- with the weak acid is assumed to run to completion & the concentrations of the acid remaining & conjugate base are determined. • Determine the major species • Since HA is stronger than H2O it is the dominant equilibrium • Then do the equilibrium (Henderson-Hasselbach)

  30. The pH Curve for the Titrations of 100.0mL of 0.050 M NH3 with 0.10 M HClweak base – strong acid At equivalence point (pH < 7): NH4+ is acidic

  31. In a titration curve • Equivalence point is defined by the stoichiometry NOT by the pH. • Remember it is where the mol of H+ are equal to the moles of OH-

  32. Summary • Strong acid and base just stoichiometry. • Weak acid with 0 ml of base - Ka • Weak acid before equivalence point • Stoichiometry first • Then Henderson-Hasselbach • Weak acid at equivalence point- Kb -Calculate concentration • Weak acid after equivalence - leftover strong base. -Calculate concentration

  33. Summary • Weak base with 0 ml of acid - Kb • Weak base before equivalence point. • Stoichiometry first • Then Henderson-Hasselbach • Weak base at equivalence point Ka. -Calculate concentration • Weak base after equivalence – left over strong acid. -Calculate concentration

  34. 15.5 Acid-Base Indicators • Two common methods for monitoring the pH • 1. pH meter • 2. acid-base indicator (this is not a good choice for finding the equivalence point.) Careful selection of indicator can help get results close. Endpoint IS change in color Equivalence point IS moles acid = moles base

  35. HIn • Most indicators are themselves a weak acid. They are one color with the proton attached and another without the proton. • Phenolphthalein, the most common indicator, is colorless as an acid and pink as a base (In-)

  36. Did the color change? • Typically, 1/10 of the initial form must be converted for the human eye to see a new color. When titrating an acidic solution…

  37. If a basic solution is titrated • Indicator will initially exist as In- and more HIn will form when acid is added

  38. Which indicator to use? • It is best to choose an indicator whose endpoint is closest to our equivalence point. • It is easier to choose an indictor if there is a large change in pH near the equivalence point (vertical area of pH curve) • The weaker the acid, the smaller the vertical area around the equivalence point, giving less flexibility in indicator choice

  39. 15.6 Solubility Equilibria and Ksp • Will it dissolve, and if not, how much? • If everything dissolves, it is an equilibrium position • Partial dissolving -- The solid will precipitate as fast as it dissolves (equilibrium) • Surface area changes the rate at which something dissolves, but NOT the amount (no change in equilibrium position)

  40. Generic equation • M+ is the cation (usually metal) • Nm- is the anion (a nonmetal) • The solubility product for each compound is found below

  41. Solubility vs. solubility product • Solubility is NOT the same as solubility product. • Solubility product is an equilibrium constant. • It doesn’t change except with temperature. • Solubility is an equilibrium position for how much can dissolve. • A common ion can change this.

  42. Let’s try a couple…first an easy one! • What is the Ksp value of copper (I) bromide with a measured solubility of 2.0 x 10-4mol/L.

  43. Now a bit more difficult • Calculate the Ksp for bismuth sulfide, which has a solubility of 1.0x10-15M at 25oC

  44. Solubility from Ksp • Find the solubility of Silver Chloride. Ksp=1.6x10-10.

  45. One more…a bit tougher • Find the solubility of copper (II) iodate, Cu(IO3)2. The Ksp for Cu(IO3)2 = 1.4x10-7 The solubility of copper (II) iodate is 3.3x10-3

  46. Comparing Ksp values • Ksp values give info about solubility (relative solubility). • When salts have the SAME number of ions: the larger the Ksp the more soluble • When salts have different number of ions: you CANNOT compare directly.

  47. Common ion effect and solubility • The presence of a common ion DECREASES the solubility of the salt. Let’s compare: What is the solubility of AgBr in both pure water & in 0.0010M NaBr. (pure H2O first)

  48. Now with a common ionWhat is the solubility of AgBrin 0.0010M NaBr

  49. Compare the two results In Water Common ion solution Notice that the common ion reduces the solubility as stated earlier (the reverse reaction occurs faster: equilibrium lies to the left)

  50. remove add pH and solubility • Presence of a common ion decreases solubility Insoluble bases dissolve in acidic solutions • Insoluble acids dissolve in basic solutions At pH less than 10.45 Lower [OH-] Increase solubility of Mg(OH)2 At pH greater than 10.45 Raise [OH-] Decrease solubility of Mg(OH)2

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