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pk 2 chem 5

# pk 2 chem 5 - Chapter 16 Do I Use the Full Equilibrium...

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Chapter 16

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2 Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation? the Henderson-Hasselbalch equation is generally good enough when the “ x is small” approximation is applicable generally, the “ x is small” approximation will work when both of the following are true: a) the initial concentrations of acid and salt are not dilute b) the K a is fairly small for most problems, this means that the initial acid and salt concentrations should be over 1000x larger than the value of K a
3 How Much Does the pH of a Buffer Change When an Acid or Base Is Added? though buffers do resist change in pH when acid or base are added to them, their pH does change calculating the new pH after adding acid or base requires breaking the problem into 2 parts 1. a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other added acid reacts with the A to make more HA added base reacts with the HA to make more A 2. an equilibrium calculation of [H 3 O + ] using the new initial values of [HA] and [A ]

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4 Ex 16.3 - What is the pH of a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L that has 0.010 mol NaOH added to it? If the added chemical is a base, write a reaction for OH with HA. If the added chemical is an acid, write a reaction for it with A . Construct a stoichiometry table for the reaction HC 2 H 3 O 2 + OH C 2 H 3 O 2 - + H 2 O HA A - OH mols Before 0.100 0.100 0 mols added - - 0.010 mols After 0.090 0.110 0
5 Ex 16.3 – Compare the effect on pH of adding 0.010 mol NaOH to a buffer that has 0.100 mol HC 2 H 3 O 2 and 0.100 mol NaC 2 H 3 O 2 in 1.00 L to adding 0.010 mol NaOH to 1.00 L of pure water? HC 2 H 3 O 2 + H 2 O C 2 H 3 O 2 - + H 3 O + + = ] [HA ] [A log p pH - a K ( 29 83 . 4 090 . 0 0.110 log 5 4 7 . 4 pH = + = p K a for HC 2 H 3 O 2 = 4.74 5 M 010 . 0 L 1.00 mol 010 . 0 ] OH [ = = - ( 29 00 . 2 10 0 . 1 log ] OH log[ pOH 2 = × - = - = - - 12.00 2.00 - 14.00 pOH - 14.00 pH 00 . 14 pOH pH = = = = +

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6 Basic Buffers B: ( aq ) + H 2 O ( l ) H:B + ( aq ) + OH ( aq ) buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B + Cl H 2 O ( l ) + NH 3 ( aq ) NH 4 + ( aq ) + OH ( aq )
7 Ex 16.4 - What is the pH of a buffer that is 0.50 M NH 3 (p K b = 4.75) and 0.20 M NH 4 Cl? find the p K a of the conjugate acid (NH 4 + ) from the given K b Assume the [B] and [HB + ] equilibrium concentrations are the same as the initial Substitute into the Henderson-Hasselbalch Equation Check the “ x is small” approximation NH 3 + H 2 O NH 4 + + OH + = + ] [HB [B] log p pH a K ( 29 ( 29 65 . 9 20 . 0

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pk 2 chem 5 - Chapter 16 Do I Use the Full Equilibrium...

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