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Acid-base equilibria & common ions Consider solution containing HF (weak acid) and salt NaF What effect does presence of NaF have on dissociation equilibrium of HF? HF(aq) H+(aq) + F-(aq) The Common Ion Effect Equilibrium calculations involving common ions 34.6 g of NH4Cl is added to 3.98 L of a 0.0145 M solution of NH3. Kb(NH3) = 1.8 x 10-5. What is the pH of the original solution before the addition of NH4Cl? NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) So – what’s the pH of the solution after the addition of the NH4Cl (assume that the volume stays constant)? Buffered solutions Buffer: A solution that resists a change in its pH when either H+ or OH- ions are added Buffers can contain weak acids & salts or weak bases & salts (solutions can be buffered at almost any pH) How does buffering work? Buffer contains relatively large concs. of weak acid HA and its conjugate base A-. Add OH- ions into solution, what happens? (OH- is a strong base – it will look for H+) Weak acid HA is best source of H+. OH- + HA H2O + A- OH- ions cannot accumulate (replaced by A- ions) Original buffered solution pH OH- added added OH- ions replaced by A- ions Final pH of buffer close to original pH How does buffering work? If OH- ions converted to A- ions, how does the pH stay so stable? Look at eq. expression for [H+]; dissociation equilibrium for HA + Ka = [H ][A ______] [HA] or [H+] = Ka[HA] ___ [A-] The essence of buffering: [HA] and [A-] are very large relative to [OH-] added. The essence of buffering Original [HA]/[A] ratio OH- added Final [HA]/[A] ratio close to original Added OH- changes HA to A-, but [HA] and [A-] are large compared to [OH-] added Buffer solution pH calculation Buffered solution contains 0.50 M HC2H3O2 (Ka = 1.8x10-5) and 0.50 M NaC2H3O2. Calculate pH of solution. HC2H3O2(aq) C2H3O2-(aq) + H+(aq) pH changes in buffered solutions Calculate change in pH that occurs when 0.010 mol solid NaOH added to 1.0 L of buffered solution from last question.Compare this with pH change that occurs when 0.010 mol solid NaOH added to 1.0 L of H2O. Expected: pH of buffered solution will change very little; pH of water will change by much larger amount. How to handle these problems Original buffered solution pH Step 1: Do stoichiometry calculations; assume reaction with H+/OH- goes to completion Modified pH Step 2: Do equilibrium calculations Adding H+, rather than OH Exactly the same thinking applied when H+ added to buffered solution of weak acid and conjugate base salt. A- (conjugate base) has high affinity for added H+, and will form weak acid HA; H+ + A- HA H+ ions cannot accumulate (replaced by HA) Net change of A- to HA, but if [A-] and [HA] are very large compared to [H+], little pH change will occur (as expected, in buffered solution). [H+] = Ka[HA] ___ [A-] Buffer Capacity Amount of H+/OH- a buffer can absorb without a considerable change in pH Buffer 1: Ka = [NH3][H+] / [NH4+] [H+] = Ka. [NH4+] / [NH3] = 5.6 x 10-10 M; pH = ? What happens to the pH if 0.1 moles of NaOH is added? After the reaction with NaOH, calculate the [H+] and thus the pH…. 1.0 M NH3, and 1.0 M NH4+ Buffer Capacity Buffer 2: 0.01 M NH3, and 0.01 M NH4+ Ka = [NH3][H+] / [NH4+] [H+] = Ka. [NH4+] / [NH3] = 5.6 x 10-10 M; pH = ? What happens to the pH if 0.1 moles of NaOH is added? After the reaction with NaOH, calculate the [H+] and thus the pH…. Henderson-Hasselbalch equation Useful equation; allows for the calculation of buffer pH if the concentration of weak acid (HA) and conjugate base (A-)are known. A more convenient method of pH calculation – will give the same answer as with our previous method of pH calculation. Buffered solution contains 0.50 M HC2H3O2 (Ka = 1.8x10-5) and 0.50 M NaC2H3O2. Calculate pH of solution (we did this a few slides ago, but let’s check that H-H equation gives us the same answer). Henderson-Hasselbalch equation Titrations / pH curves Experimental method to determine the concentration of an acid (or base). Think about this: If you were given 25 mL of a unknown concentration of HCl, a solution of 0.100 M NaOH and a burette, how could you determine the concentration of the HCl? What would happen to the pH of the solution of HCl as the NaOH was added to it? Titrations / pH curves Titration Calculations Solubility Equilibria Fluoride – used to combat tooth decay One product of F- reaction at site of teeth is CaF2(s) CaF2(s) Ca2+(aq) + 2F-(aq) (dissolving in water) Consider equlibrium set up between these species: CaF2(s) Ca2+(aq) + 2F-(aq) Equilibrium Expression for this process? Ksp = [Ca2+][F-]2 (Why is CaF2 not included in expression?) Ksp = Solubility Product Constant (solubility product) Solubility vs Ksp Ksp: Equilibrium Constant (solubility product) Solubility: Equilibrium Position Copper (I) bromide has a measured aqueous solubility of 2.0 x 10-4 mol/L at 25 °C. Calculate its Ksp value. Equilibrium reaction? Equilibrium Expression? Ksp = (equilibrium concentrations) Initial Concentrations?