BuffersBuffers

AP ChemistryAP Chemistry

Common Ion EffectCommon Ion EffectAdding a common ion to an equilibrium system Adding a common ion to an equilibrium system will shift the equilibrium forward or reverse.will shift the equilibrium forward or reverse.

For a weak acid equilibrium this can affect the For a weak acid equilibrium this can affect the pH of the solution.pH of the solution.

For example: Adding NaF to a solution of HF.For example: Adding NaF to a solution of HF.

HF HHF H++ + F + F--

Adding F- ions will shift the equilibrium to Adding F- ions will shift the equilibrium to reactants and produce more HF and reduce [H+] reactants and produce more HF and reduce [H+] ions in the process.ions in the process.

This makes the pH increase.This makes the pH increase.

BuffersBuffersBuffersBuffers are solutions in which the pH are solutions in which the pH remains relatively constant, even remains relatively constant, even when small amounts of acid or base when small amounts of acid or base are addedare added–Contain a weak acid (HA) and its Contain a weak acid (HA) and its

salt(NaA); Hsalt(NaA); H22COCO33 & NaHCO & NaHCO33

–Or a weak base (B) and its (BHCl); Or a weak base (B) and its (BHCl); (NH (NH33 and NH and NH44Cl)Cl)

BuffersBuffersA buffer system is better able to A buffer system is better able to resist changes in pH than pure waterresist changes in pH than pure waterSince it is a Since it is a pairpair of chemicals: of chemicals:–one chemical neutralizes any one chemical neutralizes any acidacid

added, while the other chemical added, while the other chemical would neutralize any additional would neutralize any additional basebase

–AND, they AND, they produce each otherproduce each other in the process!!!in the process!!!

How a Buffer WorksHow a Buffer WorksConsider the following buffer systemConsider the following buffer system

HCOHCO33-- + H + H++ H H22COCO33

If you add more HIf you add more H++ this buffer system it will this buffer system it will react with the conjugate base HCOreact with the conjugate base HCO33- to - to

produce more Hproduce more H22COCO33..

If you add base, OH-, it will grab an H+ If you add base, OH-, it will grab an H+ from Hfrom H22COCO33 to produce more HCO to produce more HCO33- ion as - ion as

followsfollows

HH22COCO33 + OH- H + OH- H22O + HCOO + HCO33--

Buffer CapacityBuffer CapacityThe The buffer capacitybuffer capacity is the amount of is the amount of acid or base that can be added acid or base that can be added before a significant change in pHbefore a significant change in pH

This depends on the amounts of HA This depends on the amounts of HA and A- present in the bufferand A- present in the buffer

Most efficient buffer is when Most efficient buffer is when

[A-][A-]

[HA][HA]= 1

Henderson-Hasselbach EquationHenderson-Hasselbach Equation

Derived from the equilibrium expression of Derived from the equilibrium expression of a weak acid and the pH equation.a weak acid and the pH equation.

This equation allows you to determine the This equation allows you to determine the pH of a buffer solutionpH of a buffer solution

pKa = -log KapKa = -log Ka

pH = pKa + log [A-][HA]

The Common Buffer ProblemsThe Common Buffer Problems1. Compute pH of a buffer given the 1. Compute pH of a buffer given the actual concentrations of the conjugate actual concentrations of the conjugate acid and conjugate base.acid and conjugate base.

The pH is easily calculated with: The pH is easily calculated with:

pH = pKa + log[A-] [HA]

Example: Determine the pH in which 1.00 Example: Determine the pH in which 1.00 mole of Hmole of H22COCO33 (Ka = 4.2 x 10 (Ka = 4.2 x 10-7-7) and 1.00 ) and 1.00

mole NaHCOmole NaHCO33 dissolved in enough water dissolved in enough water

to form 1.00 Liters of solution.to form 1.00 Liters of solution.

pH = pKa + log[A-] [HA]

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!

2. Make a buffer problem2. Make a buffer problemHow many moles of NaHCOHow many moles of NaHCO33 should be added to should be added to

1 liter of 0.100M H1 liter of 0.100M H22COCO33 (Ka = 4.2 x 10 (Ka = 4.2 x 10-7-7 ) to ) to

prepare a buffer with a pH of 7.00?prepare a buffer with a pH of 7.00?

pH = pKa + log [HCO3-][H2CO3]

7.00 = -log(4.2 x 10-7) + log [HCO3-] (0.100)

0.60 = log[HCO3-] 0.001

100.6 = [HCO3-] 0.100

[HCO3-] = 0.40 moles should be added!

3. The pH shift problem:3. The pH shift problem:

If you add 2.00 mL of 0.100 M HCl to 100 If you add 2.00 mL of 0.100 M HCl to 100 mL of a buffer consisting of 0.100 M HA mL of a buffer consisting of 0.100 M HA and 0.200 M NaA, what will be the change and 0.200 M NaA, what will be the change in pH? in pH?

Ka of HA is 1.5x10Ka of HA is 1.5x10-5-5. pKa = 4.82. . pKa = 4.82.

The initial pH is:The initial pH is:

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

mmoles H+mmoles H+ mmoles A-mmoles A-mmoles HA mmoles HA Volume, Volume,

mL mL

Initial Initial 0.200 0.200 20.0 20.0 10.0 10.0 100+2 mL 100+2 mL

Final Final 0 0 19.8 19.8 10.210.2 102 mL 102 mL

Molarities Molarities 19.8/102 19.8/102 10.2/102 10.2/102