Honors Chemistry- Buffers and Titrations

What is the role of a buffer?

Resists change in pH

Contents of a buffer?

Weak acid and it’s conjugate base

How do you make a buffer?

Combine a weak acid and it’s conj base from separate solutions

Will putting a weak acid in water work?

NO

How does a buffer acid resist pH change?

Reacts with added OH-

How does a buffer base resist pH change?

Reacts with added H+

Buffer capacity

How good a buffer is at neutralizing added H+ or OH-

Buffer better at neutralizing acid or base when HA > A-?

Better at neut added OH-, because more buff acid

Buffer better at neutralizing acid or base when HA < A- ?

Better at neut. added H+, because more buff base

What is special about the pH of a buffer?

It is close to the pKa and higher or lower based on concentrations

Henderson-Hasselbach Equation

pH = pKₐ + log([A⁻]/[HA])

Henderson-Hasselbach Equation, IMPORTANT RULE!!!!

The concentration of the conj BASE goes on top, and the concentration of an acid is on the bottom

pH and pKa relationship: (HA)=(A-)

pH=pKa

pH and pKa relationship: (A-) > (HA)

pH>pKa

pH and pKa relationship: (HA) > (A-)

pH<pKa

How to do engineering with a buffer?

Pick an acid with a pKa closest to the desired pH. For example, if you want a buffer with a pH of 5.15, the pKa is around 5, so find one raised to the -5, if not, the next best option is to go to a higher negative, so -6

What to do if the engineered buffer is too concentrated?

Multiply both top and bottom by 0.1 until a desired concentration is met

Equivalence point

the “end pt” where only products are present, the indicator is picked based off this point too

What is important about the titration problem set up?

The wording will equal the order! First one listed is in the FLASK, and second one listed is in the burette

SA+SB titration: Point A

Before the base is added, there is only strong acid in the flask

SA+SB titration: Point A: Solve for pH

Take the negative log of the concentration of (HX), by doing an ice problem and writing the chemical rxn out

SA+SB titration: Point B

Where some base is added, and in the flask there starts to form a buffer.

SA+SB titration: Point B: Solving for pH

Use stoich to find the concentration of H+ left, then take the negative log of H+

SA+SB titration: Point C

Enough base is added to reach the eq point, so there are only products left

SA+SB titration: Point C: Solving for pH

It is 7!

SA+SB titration: Point D

Base is continued to be added, there is no more strong acid left, but rather a reactant of the strong base is now there with the buffer as a product

SA+SB titration: Point D: Solving for pH

Use stoich to find the (OH-) left, and then take the -log of OH, then subtract 14 to get pH

WA+SB titration: Point A

Before adding base, there is weak acid in the flask.

WA+SB titration: Point A: Solving for pH

In order to find the pH, you must solve for Ka, and then take the -log of the H+ concentration to find the pH

WA+SB titration: Point B

Some of the base is added, which will form the weak acids conjugate base.

WA+SB titration: Point B: Solving for pH

Use stoich to find the concentration and then use the buffer equation pH = pKₐ + log([A⁻]/[HA])

WA+SB titration: Point C

Enough base is added to reach eq point, meaning there are only products left

WA+SB titration: Point C: Solving for pH

Use stoich for the concentration, and then plug that into a Kb equation, and take the -log of (OH-) and subtract 14 to get your pH

WA+SB titration: Point D

Base contiunes to be added, leaving the strong base as a reactant, and products are intact

WA+SB titration: Point D: Solving for pH

Stoich for the concentration of (OH-), and take the -log of (OH-) and subtract 14 to get pH

SA+WB titration: Point A

Before acid is added, there is still weak base in the flask

SA+WB titration: Point A: Solving for pH

Find your Kb, and then take the -log of (OH-) and subtract 14 to get pH

SA+WB titration: Point B

Some of the weak acid is added, which starts to form the products and the buffer

SA+WB titration: Point B: Solving for pH

Stoich to find concentration, and then use the buffer equation to find pH pH = pKₐ + log([A⁻]/[HA])

SA+WB titration: Point C

Enough acid is added to reach the eq point, so there are only products left

SA+WB titration: Point C: Solving for pH

Stoich to find the concentration, and then find the Ka and use that H+ concentration to take the -log of it to obtain pH

SA+WB titration: Point D

Acid will continue to be added, leaving only the strong acid as a reactant and the product

SA+WB titration: Point D: Solving for pH

Use stoich to find the H+ concentration and take the -log of H+

Under the eq point, what does it establish on the x-axis?

X amount of titrant added

Under the point ½ less than the eq point, what does it establish on the x-axis, and what can we call this point?

½ x. the ½ equiv. volume

What is special at the ½ equiv. volume point?

pH=pKa, where (HA)=(A-)

If the equivlance point is greater than 7, there is more…

Buffer base

NEVER ASSUME THAT THE EQ POINT IS

7

What two points can the ½ eq point form?

It can be in between point X and point Y, where point Y is always closer to the eq point, this is known as the buffer zone

On an ascending pH graph, what is the buffer zone relationship between X and Y?

X: pH<pKa

y: pH>pKa

On an descending pH graph, what is the buffer zone relationship between X and Y?

X: pH>pKa

Y: pH<pKa