1. Chapter 16 Aqueous Ionic Equilibrium: 16.3-16.4
By Michael Kahn

2. Factors that influence the effectiveness of a buffer
An effective buffer neutralizes small to moderate amounts of added acid or base. A buffer can be destroyed by adding too much acid or too much base
Factors that influence effectiveness
The relative amounts of the acid and conjugate base
The absolute concentrations of the acid and conjugate base

3. Factors that influence the effectiveness of a buffer
Relative amounts of Acid and Base
A buffer is most effective (most resistant to pH changes) when the concentrations of acid and conjugate base are equal. When the conjugate acid and base components are equal, the pH is equal to the pKa of the weak acid of the buffer system
When more of the acid component is present, the pH becomes more acidic. When more of the base component is present, the pH becomes more basic. The pH in both cases can be calculated using the Henderson-Hasselbalch equation

4. Absolute concentrations of the Acid and Conjugate Base
When the concentration of the conjugate acid and base components of a buffer system are equal, the pH is equal to the pKa of the weak acid of the buffer system.
Acid added: When a small amount of a strong acid is added, it reacts with the conjugate base of the buffer system, converting it to the weak acid of the buffer system. Thus, the weak acid concentration increases , the conjugate base concentration decreases and the pH drops slightly.
Base added: When a small amount of a strong base is added, it reacts with the weak acid of the buffer system, converting it to the conjugate base of the buffer system. Thus, the weak acid concentration decreases, the conjugate base concentration increases, and the pH rises slightly

5. Finding pH of solution
1)determine which component is the acid, determine which component is the base, and substitute their concentrations in to the Henderson-Hasselbalch equation.
2)The pKa is the negative of the log of the equilibrium constant of the acid dissociation reaction where the weak acid component and water are the reactants and the conjugate base and h30+ are the products.
3)Lastly, confirm that the “x is small” approximation is valid by calculating the [h3o+] from the ph.
Because h3o+ is formed by ionization of the acid, the calculated [h3o+] has to be less than .05/5% of the initial concentration of the acid for the “x is small” approximation to be valid

6. Factors that influence the effectiveness of a buffer
The factors that influence the effectiveness of a buffer are the relative amounts of the acid and conjugate base
Relationships
(the closer they are to each other, the more effective the buffer)
the absolute concentrations of the acid and base conjugate base (the higher the absolute concentrations, the more effective the buffer).

7. Effectiveness of buffer
The relative concentrations of acid and conjugate base should not differ by more than a factor of 10 for a buffer to be reasonably effective.
Using the Henderson-Hasselbalch equation, this means that the pH should be within one pH unit of the weak acid’s pKa

8. Question
Which acid would you choose to combine with its sodium salt to make a solution buffered at pH 4.25? For the best choice, calculate the ratio of the conjugate base to the acid required to attain the desired pH?
A)chlorous acid (HClO2) pKa= 1.95
B)nitrous acid (HNO2) pKa= 3.34
C) formic acid (HCHO2) pKa= 3.74
D)hypochlorous acid (HClO) pKa= 7.54

9. Solution Answer: formic acid
Reason: its pKa lies closest to the desired pH.
Proof:
pH= pKa+log([base]/[acid])
4.25=3.74+log([base]/[acid])
Log([base]/[acid])= =.51
([base]/[acid])=10^0.51
=3.24

10. Acid-Base Titration
In an acid-base titration, a basic/acidic solution of unknown concentration is reacted with an acidic/basic solution of known concentration. The known solution is slowly added to the unknown solution while the pH is monitored with either a pH meter or an indicator (a substance whose color depends on pH).
As the acid and base combine, they neutralize each other. At the equivalence point- the point in the titration when the number of moles of base is stoichiometrically equal to the number of moles of acid –the titration is complete.
When this point is reached, neither reactant is in excess and the number of moles of the reactants are related by the reaction stoichiometry

11. Titrations
The titration of weak acid by a strong base always has a basic equivalence point because at the equivalence point, all of the acid has been converted into its conjugate base, resulting in a weakly basic solution
The volume required to get to the equivalence point is only dependent on the concentration and volume of acid or base to be titrated and the base or acid used to do the titration because the equivalence point is dependent on the stoichiometry of the balanced reaction of the acid and base.
The stoichiometry only considers the number of moles involved, not the strength of the reactants involved

12. The Titration of a Strong Acid with a Strong Base
1) the initial pH of the solution is simply the pH of the strong acid. Because strong acids completely dissociate the concentration of H3O+ is the concentration of the strong acid and pH=-log[H3O+]
2) before the equivalence point, H3O+ is in excess . Calculate the [H3O+] by subtracting the number of moles of added OH- from the initial number of moles of H3O+ and dividing by the total volume. Then convert to pH using –log [H3O+]
3) At the equivalence point, neither reactant is in excess and the pH=7.00 (neutral)
4) beyond the equivalence point, OH- is in excess. Calculate the [OH-] by subtracting the initial number of moles of H3O+ from the number of moles of added OH- and dividing by the total volume. Then convert to pH using –log[H3O+]

13. Example
A 50.0 ml sample of .200 M sodium hydroxide is titrated with .200 M nitric acid. Calculate pH.
A) after adding 30 ml of HNO3
B) at the equivalence point

14. solution A)
1) Moles NaOH=.0500Lx(.200 mol)/(1L)=.0100mol NaOH strong base so dissociates completely so the amount of OH- is equal to the amount of NaOH
moles OH-=.0100 mol
2) Mol HNO3 added=.0300Lx(.200mol)/(1L)calculate the amount of moles added at 30.0 mL from molarity
=.006molHNO3
) OH ) h30 )  h2o
Before addition) mol ) 0.0mol )  calculate the number of moles of OH- by setting up a table based on
Addition ) ) mol ) the neutralization reaction that shows the amount of OH- before the
After addition ) mol ) mol ) addition, amount of H30+ added, and the amounts left after addition
3) [OH-]=(.0040mol)/(.05L + .03L)= .05M  calculate the OH- concentration by dividing amount of OH-
remaining by the total volume-(initial plus added volume)
4)pOH= -log(.0.05)
=1.30
5)pH=14-pOH =
=12.70 B)
6)pH= at the equivalence point, the strong base has completely
neutralized the strong acid (pH neutral)

15. College board course overview
LO 6.13 e student can interpret titration data for monoprotic or polyprotic acids involving titration of a weak or strong acid by a strong base (or a weak or strong base by a strong acid) to determine the concentration of the titrant and the for a weak acid, or the for a weak base. [See SP 5.1, 6.4, connects to 1.E.2]
LO 6.14 e student can, based on the dependence of on temperature, reason that neutrality requires as opposed to requiring , including especially the applications to biological systems. [See SP 2.2, 6.2]
LO 6.18 the student can design a buffer solution with a target pH and buffer capacity by selecting an appropriate conjugate acid-base pair and estimating the concentrations needed to achieve the desired capacity. [See SP 2.3, 4.2, 6.4]
LO 6.20 e student can identify a solution as being a buffer solution and explain the buffer mechanism in terms of the reactions that would occur on addition of acid or base. [See SP 6.4]