Click here to load reader

Aqueous Equilibria Chapter 15 Applications of Aqueous Equilibria

  • View
    235

  • Download
    1

Embed Size (px)

Text of Aqueous Equilibria Chapter 15 Applications of Aqueous Equilibria

  • Slide 1

Aqueous Equilibria Chapter 15 Applications of Aqueous Equilibria Slide 2 Aqueous Equilibria THE COMMON-ION EFFECT Consider a solution of acetic acid: If acetate ion is added to the solution, Le Chtelier says the equilibrium will shift to the left. HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 (aq) Slide 3 Aqueous Equilibria THE COMMON-ION EFFECT The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte. Add more of the same ion and there will be less ions of the weak one. Slide 4 Aqueous Equilibria THE COMMON-ION EFFECT The addition of concentrated HCl to a saturated solution of NaCl will cause some solid NaCl to precipitate out of solution. The NaCl has become less soluble because the addition of additional chloride ion. NaCl + HCl more NaCl due to increased Cl - Slide 5 Aqueous Equilibria THE COMMON-ION EFFECT pH If a substance has a basic anion, it will be more soluble in an acidic solution. Substances with acidic cations are more soluble in basic solutions. Slide 6 Aqueous Equilibria THE COMMON-ION EFFECT The addition of a common ion to a weak acid solution makes the solution LESS acidic. HC 2 H 3 O 2(aq) H + (aq) + C 2 H 3 O 2 - (aq) If NaC 2 H 3 O 2 is added to the system, the equilibrium shifts to undissociated HC 2 H 3 O 2 raising the pH. The new pH can be calculated by putting the concentration of the anion into the K a equation and solving for the new [H + ]. Slide 7 Aqueous Equilibria THE COMMON-ION EFFECT Adding NaF to a solution of HF causes more HF to be produced. Major species are HF, Na +, F -, and H 2 O. Common ion is F -. In a 1.0M NaF and 1.0M HF solution, there is more HF in the presence of NaF. HF (aq) H + (aq) + F - (aq) Le Chateliers indicates that additional F - due to the NaF causes a shift to the left and thus generates more HF. Slide 8 Aqueous Equilibria THE COMMON-ION EFFECT Finding the pH 1.Always determine the major species. 2.Write the equilibrium equation and expression. 3.Determine the initial concentrations. 4.Do ICE chart and solve for x. 5.Once [H + ] has been found, find pH. Slide 9 Aqueous Equilibria PRACTICE ONE The equilibrium concentration of H + in a 1.0M HF solution is 2.7 x 10 -2 M and the percent dissociation of HF is 2.7%. Calculate [H + ] and the percent dissociation of HF in a solution containing 1.0M HF (K a = 7.2 x 10 -4 ) and 1.0M NaF. Slide 10 Aqueous Equilibria THE COMMON-ION EFFECT Notice: 1.0M HF% dissociation 2.7% versus 1.0M HF and 1.0M NaF% dissociation 0.072% Slide 11 Aqueous Equilibria The Common-Ion Effect Because HCl, a strong acid, is also present, the initial [H 3 O + ] is not 0, but rather 0.10 M. [HF], M[H 3 O + ], M[F ], M Initially0.200.100 Changexx+x+x+x+x At Equilibrium 0.20 x 0.200.10 + x 0.10 x HF (aq) + H 2 O (l) H 3 O + (aq) + F (aq) Slide 12 Aqueous Equilibria EQUATIONS QUIZ For each of the following reactions, write an equation for the reaction. Write the net ionic equation for each. Omit formulas for spectator ions or molecules in the reaction. Put a box around your final answer. Slide 13 Aqueous Equilibria DO NOW Pick handout due tomorrow. Turn in lab make sure the you have: One Title page, one Prelab, one Data Table, Calculations Table for everyone, and Calculations 1-8 for everyone in that order! Get out notes. Slide 14 Aqueous Equilibria BUFFERS Solutions of a weak conjugate acid-base pair. They are particularly resistant to pH changes, even when strong acid or base is added. Just a case of the common ion effect. Slide 15 Aqueous Equilibria BUFFERS You are always adding a strong acid or strong base to a buffer solution. Buffer system (conjugate acid-base pair) acts as a net catches acid or base Slide 16 Aqueous Equilibria BUFFERS If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH to make F and water. Slide 17 Aqueous Equilibria BUFFERS If acid is added, the F reacts to form HF and water. Slide 18 Aqueous Equilibria BUFFERS Example: HC 2 H 3 O 2 / C 2 H 3 O 2 - buffer system (the net) Add a strong acid: H + + C 2 H 3 O 2 - HC 2 H 3 O 2 forms a weak acid Add a strong base: OH - + HC 2 H 3 O 2 C 2 H 3 O 2 - + H 2 O forms a weak base Slide 19 Aqueous Equilibria BUFFERS Example: NH 3 / NH 4 + buffer system (the net) Add a strong acid: H + + NH 3 NH 4 + forms a weak acid Add a strong base: OH - + NH 4 + NH 3 + H 2 O forms a weak base Slide 20 Aqueous Equilibria PRACTICE TWO A buffered solution contains 0.050M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) and 0.50M sodium acetate (NaC 2 H 3 O 2 ). Calculate the pH of the solution. Slide 21 Aqueous Equilibria HELPFUL TIPS Buffered solutions are simply solutions of weak acids and bases containing a common ion. The pH calculations for buffered solutions require exactly the same procedures as determining the pH of weak acid or weak base solutions learned previously. When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. After the stoichiometric calculations are completed, then consider the equilibrium calculations. Slide 22 Aqueous Equilibria BUFFERS Adding a strong acid or base to a buffered solution Requires moles Slide 23 Aqueous Equilibria PRACTICE THREE Calculate the change in pH that occurs when 0.010mol solid NaOH is added to 1.0L of the buffered solution contains 0.050M acetic acid (HC 2 H 3 O 2, K a = 1.8 x 10 -5 ) and 0.50M sodium acetate (NaC 2 H 3 O 2 ). Compare this pH change with that which occurs when 0.010mol solid NaOH is added to 1.0L of water. Slide 24 Aqueous Equilibria HOW BUFFERING WORKS Slide 25 Aqueous Equilibria HOW BUFFERING WORKS Slide 26 Aqueous Equilibria HOW BUFFERING WORKS So [H + ] (and thus pH) is determined by the ratio of [HA]/[A - ]. When OH - ions are added, HA converts to A - and the ratio decreases. However, is the amounts of [HA] and [A - ] are LARGE, then the change in the ratio will be small. If [HA]/[A - ] = 0.50M / 0.50M = 1.0M initially. After adding 0.010M OH -, it becomes [[HA]/[A - ] = 0.49M / 0.51M = 0.96M. Not much of a change. [H + ] and pH are essentially constant. Slide 27 Aqueous Equilibria HOW BUFFERING WORKS Slide 28 Aqueous Equilibria BUFFER CALCULATIONS Consider the equilibrium constant expression for the dissociation of a generic acid, HA: [H 3 O + ] [A ] [HA] K a = HA + H 2 OH 3 O + + A Slide 29 Aqueous Equilibria BUFFER CALCULATIONS Rearranging slightly, this becomes [A ] [HA] K a = [H 3 O + ] Taking the negative log of both sides, we get [A ] [HA] log K a = log [H 3 O + ] + log pKapKa pH acid base Slide 30 Aqueous Equilibria BUFFER CALCULATIONS So pK a = pH log [base] [acid] Rearranging, this becomes pH = pK a + log [base] [acid] This is the HendersonHasselbalch equation. Slide 31 Aqueous Equilibria BUFFER CALCULATIONS For a particular buffering system, all solutions that have the same ratio of [A - ] /[HA] have the same pH. Optimum buffering occurs when [HA] = [A - ] and the pK a of the weak acid used should be as close to possible to the desired pH of the buffer system. Slide 32 Aqueous Equilibria HENDERSON-HASSELBACH The equation needs to be used cautiously. It is sometimes used as a quick, easy equation in which to plug in numbers. A K a or K b problem requires a greater understanding of the factors involved and can ALWAYS be used instead of the HH equation. However, at the halfway point (as in a titration), the HH is very useful. Slide 33 Aqueous Equilibria PRACTICE FOUR What is the pH of a buffer that is 0.75 M lactic acid, HC 3 H 5 O 3, and 0.25 M in sodium lactate? K a for lactic acid is 1.4 10 4. Slide 34 Aqueous Equilibria HENDERSONHASSELBALCH EQUATION pH = pK a + log [base] [acid] pH = log (1.4 10 4 ) + log (0.25) (0.75) pH = 3.85 + (.477) pH = 3.37 Slide 35 Aqueous Equilibria HINTS 1.Determine the major species involved. 2.If a chemical reaction occurs, write the equation and solve stoichiometry. 3.Write the EQ equation. 4.Set up the equilibrium expression (K a or K b ) of the HH equation. 5.Solve. 6.Check the logic of the answer. Slide 36 Aqueous Equilibria PRACTICE FIVE A buffered solution contains 0.25M NH 3 (K b = 1.8 x 10 -5 ) and 0.40M NH 4 Cl. Calculate the pH of this solution. Slide 37 Aqueous Equilibria PRACTICE SIX Calculate the pH of the solution that results when 0.10mol gaseous HCl is added to 1.0Lof the buffered solution of contains 0.25M NH 3 (K b = 1.8 x 10 -5 ) and 0.40M NH 4 Cl. Slide 38 Aqueous Equilibria BUFFERING CAPACITY This is the amount of acid or base that can be absorbed by a buffer system without a significant change in pH. In order to have a large buffer capacity, a solution should have large concentrations of both buffer components. Slide 39 Aqueous Equilibria PRACTICE SEVEN Calculate the change in pH that occurs when 0.010mol gaseous HCl is added to 1.0L of each of the following solutions (K a for acetic acid = 1.8 x 10 -5 ): Solution A: 5.00M HC 2 H 3 O 2 and 5.00M NaC 2 H 3 O 2 Solution B: 0.050M HC 2 H 3 O 2 and 0.050M NaC 2 H 3 O 2 Slide 40 Aqueous Equilibria HINT We see that the pH of a buffered solution depends on the ratio of the [base] to [acid] (or [acid] to [base]). Big concentration difference = large pH change Slide 41 Aqueous Equilibria PRACTICE EIGHT A chemist needs a solution buffered at pH 4.30 and can choose from the following list of acids and their soluble salt: a. chloroacetic acid Ka = 1.35 10 -3 b. propanoic acid Ka = 1.3 10 -5 c. benzoic acid Ka = 6.4 10 -5 d. hypochlorus acid Ka = 3.5 10 -8 Calculate the ratio of [HA] / [A - ] required for each system to yield a pH of 4.30. Which system works best? Slide 42 Aqueous Equilibria TITRATIONS and pH CURVES Only when the acid AND base are both strong is the pH at the equivalence point 7. Any other conditions and you get to do an equilibrium problem. It is really a

Search related