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Buffers

Buffers . AP Chemistry. Common Ion Effect. Adding a common ion to an equilibrium system will shift the equilibrium forward or reverse. For a weak acid equilibrium this can affect the pH of the solution. For example: Adding NaF to a solution of HF. HF H + + F -

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Buffers

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  1. Buffers AP Chemistry

  2. Common Ion Effect • Adding a common ion to an equilibrium system will shift the equilibrium forward or reverse. • For a weak acid equilibrium this can affect the pH of the solution. • For example: Adding NaF to a solution of HF. • HF H+ + F- • Adding F- ions will shift the equilibrium to reactants and produce more HF and reduce [H+] ions in the process. • This makes the pH increase.

  3. Buffers • Buffers are solutions in which the pH remains relatively constant, even when small amounts of acid or base are added • Contain a weak acid (HA) and its salt(NaA); H2CO3 & NaHCO3 • Or a weak base (B) and its (BHCl); (NH3 and NH4Cl)

  4. Buffers • A buffer system is better able to resist changes in pH than pure water • Since it is a pair of chemicals: • one chemical neutralizes any acid added, while the other chemical would neutralize any additional base • AND, they produce each other in the process!!!

  5. How a Buffer Works • Consider the following buffer system • HCO3- + H+H2CO3 • If you add more H+ this buffer system it will react with the conjugate base HCO3- to produce more H2CO3. • If you add base, OH-, it will grab an H+ from H2CO3 to produce more HCO3- ion as follows • H2CO3 + OH- H2O + HCO3-

  6. Buffer Capacity • The buffer capacity is the amount of acid or base that can be added before a significant change in pH • This depends on the amounts of HA and A- present in the buffer • Most efficient buffer is when [A-] [HA] = 1

  7. Henderson-Hasselbach Equation • Derived from the equilibrium expression of a weak acid and the pH equation. • This equation allows you to determine the pH of a buffer solution • pKa = -log Ka [A-] [HA] pH = pKa + log

  8. The Common Buffer Problems • 1. Compute pH of a buffer given the actual concentrations of the conjugate acid and conjugate base. • The pH is easily calculated with: [A-] [HA] pH = pKa + log

  9. Example: Determine the pH in which 1.00 mole of H2CO3 (Ka = 4.2 x 10-7) and 1.00 mole NaHCO3 dissolved in enough water to form 1.00 Liters of solution. [A-] [HA] pH = pKa + log pH = -log (4.2 x 10-7) + log (1.00M)/(1.00M) pH = pKa = 6.4 ** pKa of a weak acid can help determine pH of buffer you will make if it is mixed in a 1:1 mole ratio!

  10. 2. Make a buffer problem • How many moles of NaHCO3 should be added to 1 liter of 0.100M H2CO3 (Ka = 4.2 x 10-7 ) to prepare a buffer with a pH of 7.00? [HCO3-] [H2CO3] pH = pKa + log 7.00 = -log(4.2 x 10-7) + log [HCO3-] (0.100) 0.60 = log[HCO3-] 0.001 [HCO3-] = 0.40 moles should be added! 100.6 = [HCO3-] 0.100

  11. 3. The pH shift problem: • If you add 2.00 mL of 0.100 M HCl to 100 mL of a buffer consisting of 0.100 M HA and 0.200 M NaA, what will be the change in pH? • Ka of HA is 1.5x10-5. pKa = 4.82. • The initial pH is:

  12. If we add H+ to the equilibrium system it will react with the strong conjugate base ‘A-, tor form more HA. • H+ + A- HA • Set up an ICE chart to determine how the mole values will change using stoichiometry!

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