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Applications of aqueous equilibria

Applications of aqueous equilibria. Neutralization Common-Ion effect Buffers Titration curves Solubility and K sp. Learning objectives. Determine extent of neutralization Construct buffer solutions Derive Henderson-Hasselbalch equation Apply HH to calculate pH of buffer solutions

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Applications of aqueous equilibria

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  1. Applications of aqueous equilibria Neutralization Common-Ion effect Buffers Titration curves Solubility and Ksp

  2. Learning objectives • Determine extent of neutralization • Construct buffer solutions • Derive Henderson-Hasselbalch equation • Apply HH to calculate pH of buffer solutions • Calculate pH titration curves • Write solubility product expressions • Identify factors that affect solubility

  3. Neutralization • Acid + Base → Salt + Water • Extent → depends on type of acid and base • What are the major species left when base neutralizes an acid? • Four combinations: • Strong-strong • Strong-weak • Weak-strong • Weak-weak

  4. Strong-strong • Net ionic equation • Kn = 1/Kw = 1014 • Neutralization goes to completion • Na+, Cl- and H2O

  5. Manipulating equilibria expressions • If reaction A can be written as the sum of reactions B + C + D +...+ N (ΣRi) • Then K is product over ΠKi • This approach is used routinely in solution equilibria problems

  6. Weak-strong • Acetic acid is not completely ionized • Net ionic equation: • What is K? Write Kin terms equilbria we know: • K= 1.8x10-5x1.0x1014 = 1.8x109 • Goes to completion – attraction of OH- for protons • Na+, Ac-, H2O (v. small amount OH-)

  7. Strong-weak • Net ionic equation for neutralization of ammonia with HCl • What is K? Write in terms of known equilibria: • Add 1 and 2: • K1+2 = K1K2 = 1.8x10-5x1.0x1014 = 1.8x109 • K1+2>>1 - Neutralization complete • NH4+, Cl-, H2O and v. small amount of H3O+

  8. Weak-weak • Net ionic equation for neutralization of acetic acid by ammonia – neither ionized • Obtain Kn from equilibria we know • K1+2+3 = K1K2K3 • = 1.8x10-5x1.8x10-5x1.0x1014 = 3.2x104 • NH4+, Ac-, H2O (v. small amounts of HAc and NH3)

  9. Common ion effect - buffering • Solutions of weak acid and conjugate bases have important applications for “buffering” pH – resisting change to pH from added acid or base. (weak base and conjugate acid perform the same function) • The pH of operation will depend on the dissociation constants for the particular acid (base).

  10. Use weak acid strategy for calculating pH in buffer • Acetic acid and sodium acetate 0.1 M. • Initial species are HAc, Na+, Ac- and H2O • Two proton exchange reactions, but one does not alter concentrations

  11. The Big Table of concentrations • Determination of final concentrations in terms of initial concentrations • Note difference between acid case and buffer case: • [Ac-]i is 0 with HAc only • [Ac-]i is > 0 in the buffer

  12. Solving for x • Put concentrations into expression for Ka • 0.10 – x ≈ 0.10 ≈ 0.10 + x • x = Ka = 1.8 x10-5 M (life is good) • pH = 4.74 • Note: when [HAc] = [Ac-], pH = pKa • In all buffers pH = pKa when [HB] = [B-] X << 0.1

  13. Common-ion effect – Le Chatelier in action • Without added acetate ion the pH of 0.10 M acetic acid is 2.89. • Addition of Ac- causes [H3O+] to decrease • Consequence of Le Chatelier: increasing [product] equilibrium goes to reactants • pH changes in different buffers

  14. Modulating pH in acetic acid by adding acetate ion

  15. Do buffers really work? • Compare effect of adding OH- to solution of given pH • Buffer solution • Strong acid at same initial pH

  16. Adding base to buffer • Analytical proof in HAc/Ac- system • Initial pH = 4.74 • What happens when 0.01 mol of NaOH is added to I L

  17. What happens when 0.01 mol of NaOH is added? • Addition of OH- causes conversion of HAc into Ac-

  18. After addition of OH-, recompute [H3O+] • [H3O+] = 1.8x10-5x0.09/0.11 = 1.5x10-5 M • pH = 4.82 – a change of + 0.08 pH units

  19. Same deal when adding acid • Almost all the added H+ is converted into HAc • In same way as with base, new [HAc] = 0.11 M and new [Ac-] = 0.09 M • [H3O+] = 1.8x10-5x0.11/0.09 = 2.2 x 10-5 M • pH = 4.66 – a change of -0.08 pH units

  20. Adding OH- to strong acid at pH = 4.74 • What is change in pH if added to a solution of HCl at pH 4.74? • [H+] = 1.8 x 10-5 M • HCl removes 1.8x10-5 mol of OH-. • Excess OH- = 0.01 - 1.8x10-5 mol of OH- ≈ 0.01 mol. • New pH = 12 – a change of 7 pH units • In buffer at same initial pH, change was only 0.08 units

  21. Buffer capacity • A measure of the amount of acid or base that a buffer solution can absorb before a significant change in pH occurs. • Or – the amount of acid or base required to yield a given change in pH. • Messin’ with buffers

  22. Henderson-Hasselbalch:Send the pain below • Buffer calculation short-cut • Derivation • Assumes that [H+] << [HA]

  23. Applications • Able to predict percent dissociation of an acid from the difference between pH and pKa • Calculate relative quantities of acid and conjugate base required to achieve a given pH (pKa must be reasonably close to pH)

  24. The glass half-neutered • Instead of using a source of weak acid and a conjugate base, take a weak acid and neutralize with strong base to make the conjugate base. (HAc and NaOH). • When half the HAc is neutralized [HAc] = [Ac-] pKa = pH • Useful buffers can be made in pH range of pKa ± 1 pH units

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