# Chapter 16: Aqueous Ionic Equilibria Common Ion Effect Buffer Solutions Titrations Solubility Precipitation Complex Ion Equilibria.

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Chapter 16: Aqueous Ionic Equilibria Common Ion Effect Buffer Solutions Titrations Solubility Precipitation Complex Ion Equilibria

Acid Base Reactions Neutralization Reactions Strong base + Strong acid Reacts completely (neutral at eq. pt.) Strong base + Weak acid All base reacts (basic at eq. pt.) Weak base + Strong acid All acid reacts (acidic at eq. pt.) Equivalence point: equal moles acid and base have reacted

Common Ion Effect HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (aq) Adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left Therefore, the pH will be higher than the pH of the acid solution lowering the H 3 O + ion concentration

Calculate the pH of a solution which is 1.0 M HF and 1.0M NaF. Note: Last chapter we calculated pH of 1.0 M solution of HF only. We calculated the pH to be 1.57.

Buffers Buffers: Solutions that resist changes in pH when a small amount of acid or base is added. They act by neutralizing the added acid or base Many buffers are made by mixing:  A weak acid with its conjugate base anion HF + NaF   acidic buffer  A weak base with it’s conjugate acid cation NH 3 + NH 4 +   basic buffer

How Acid Buffers Work HA (aq) + H 2 O (l)  A − (aq) + H 3 O + (aq) Acid species (HA) reacts with added base Base species (A − ) reacts with added acid Therefore, small amounts of acid or base are neutralized

H2OH2O How Buffers Work HA  + H3O+H3O+ A−A− A−A− Added H 3 O + new HA

H2OH2O How Buffers Work HA  + H3O+H3O+ A−A− Added HO − new A − A−A−

Calculate the pH of a buffer which is 0.50M in acetic acid (CH 3 CO 2 H) and 0.50M in sodium acetate (NaCH 3 CO 2 ). Calculate the pH after 0.01 mol HCl is added to one liter of the above buffer. Compare with the pH of one liter of water after 0.01 mol of HCl is added to it.

Henderson-Hasselbalch Equation Calculating the pH of a buffer solution can be simplified by using an equation derived from the K a expression called the Henderson- Hasselbalch Equation The equation calculates the pH of a buffer from the K a and initial concentrations of the weak acid and salt of the conjugate base Note: the “x is small” approximation must be valid

Deriving the Henderson-Hasselbalch Equation ()()()()()()

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” (conjugate acid/base initial concentrations high, K a small) Rule of thumb: the initial acid and salt concentrations should be over 1000x larger than the value of K a

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)

The Henderson-Hasselbalch Equation for Basic Buffers

Calculate the pH of a buffer solution containing 0.25 M NH 3 and 0.40 M NH 4 Cl. K b (NH 3 )= 1.8 x10 -5 Then, if we add 20mL of 1.0M NaOH to 500mL of the above buffer, what would the final pH be? Note: the volume of the buffer changes in this problem, so we need to work in moles and calculate new concentrations.

We can predict whether a buffer will be acidic or basic by comparing values of Ka and Kb. Acetic acid CH 3 CO 2 HK a =1.8 x 10 -5 Acetate ion CH 3 CO 2 - K b = 5.6 x 10 -10 K a > K b so buffer will be acidic Ammonia NH 3 K b =1.8 x 10 -5 Ammonium ion NH 4 + K a = 5.6 x 10 -10 K b > K a so buffer will be basic

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