Buffers Flashcards
Buffers
-buffers are solutions that resist pH changes when an acid or base is added
-A buffer contains significant amounts of both a weak acid and its conjugate base or a weak base and it’s conjugate acid
Ex: CH3COOH/CH3COO
How does a buffer work?
-A buffer contains HA and A-
-when an acid is added, H+ is neutralized by the base A-
-when a base is added, OH is neutralized by the acid HA
-single arrows are used here
-A buffer contains a conjugate acid base pair, which which is chemical equilibrium
HA+H20 producing A- + H3O
Or
B+ + H2O producing BH+ + OH
How does a buffer work?
-when an acid is added, added H3O is consumed by reacting with some A- to HA
-when a base is added, OH- from the strong base is consumed by reacting with some HA to A-
-ph doesn’t change significantly because [H+] stays relatively constant
Conditions for a buffer
-to be a buffer, it must satisfy
- It contains both a weak acid and its conjugate base or a weak base and it’s conjugate acid, the weak base cannot be water
- It contains significant amount of acid/base pair
-A buffer stop buffering pH when HA or A- is used up
Henderson-Hasselbalch equation
-A buffer solution contains HA and A-
pH= pka + log [moles of A-]/[moles ofHA]
For a base: pka + log [moles of B]/moles of [BH+]
To find buffer pH
- write Henderson-Hasselbalch equation
- Determine acid and base
- Find pka and mols of acid and base
- Plug into equation
Find the amount of HA and A- needed to make a desired buffer
- Find pka
- Find moles of acid or base
- 10^(ph + pka)
- Multiply known moles with the above answer and then solve for mass
Find the amount of strong acid/base needed to mix with A-/HA to make a buffer
-single arrows are used because a strong acid/base reacts with a weak base/acid essentially completely because the equilibrium constant is large
- Calculate moles of strong acid and base
- Determined reaction change (subtract mols)
- Set up ICF table to find final mol value
ICE TABLE & ICF TABLE
-ICE table: use when K (equilibrium constant is involved in calculations)
Calculating pH changes in a buffer
-buffers are solutions that resist changes in pH when an acid or base is added, this does not mean the pH stays constant instead pH changes by a small amount to an acid or base is added to a buffer
If strong base is added to the buffer : HA decreases, and A- increases, causing pH to increase with the addition of strong base
If strong acid is added A- decreases and HA increases. Thus pH decreases with the addition of strong acid.
A good buffer
-A good buffer should be able to neutralize larger amounts of strong acid/base before its Ph changes substantially
-the effectiveness of a buffer solution is determined by two factor
- The larger the concentration/amount of the acid and conjugate base the more effective the buffer resists pH since it can neutralize larger amounts of acid or base
- The relative amount of acid and its conjugate base, a buffer is most effective when the concentration of the acid and conjugate base are equal
-an effective buffer range is between pka -1 and pka +1, buffer is most effective when ph = pka, therefore a select an acid whose pka is as close to its desired pH
Titration
-technique where a solution of known concentration is used to determine the concentration of an unknown solution
-typically the Titrant [known solution] is added from a buret to unknown quantity of the analyte [unknown solution] until the reaction is complete
Acid base titration
A basic (or acidic) solution of unknown concentration is reacted with an acidic (or basic) solution with known concentration until reaction is complete
Titration is monitored by a pH metre or a pH indicator (colour change)
Titration of a weak acid with strong base
Ex: HF is titrated with NaOH
1 region: Before titration begins (no base added)
-ph determined from weak acid equilibrium (ICE TABLE)
2 region. Before nOH<nHA
-there is a mixture of unreacted HA+A- produced by titration reaction, which makes a buffer, this region is a buffer region, whose pH barely changes with the addition of strong base/acid
-ph is calculated from Henderson Hasalbach equation (ICF table)
- At equivalent point, just enough OH- has been added to consume HA and A-
Ph is determined by Kb=kw/ka
[A-] is now moles of HA/total volume
-the ph at this region is always >7
Region 4: after equivalent point
Ph is determined by excess strong base in solution
-set up ICF table with HA+NAOH producing A- + H2O
Solubility products
-equilibrium constant for the disassociation of a solid salt into its aqueous ions: Ksp = [M^m+]^n [X^n-]^m
solids are omitted in this expression
Solubility
-amount of solute that will dissolve in a given amount of solution at a particular temperature
-Molar solubility: number of moles of solute (precipitate) that dissolve in a litre of solution in mol/L
-to compare Ksp the compounds must have the same disassociation stoichiometry
Parameters that affect solubility
-molar solubility can change at different conditions
-solubility decreases if the equilibrium is shifted to the left
-solubility increases if the equilibrium is shifted to the right
Changes in molar solubility
- If common ions are added, the equilibrium shifts to the left decreasing solubility
Ex: adding NaF (which provides F- ions) reduces solubility because they are already F- ions present in solution, but Ksp does not change
- If a salt contains a basic anion like OH- CO3 F- S, decreasing ph
increases its solubility - Neutral cations and anions are not affected by pH change (BaSO4, CaSO4)
Complex ions
-ions that formed by combining a cat ion with several and ions or neutral molecules
Ex: Ag(NH3)2
-the attached ions or molecules are called ligands (NH3)
Complex ion equilibrium
-the reaction between an ion and ligands form a complex ion is called a complex ion formation reaction
-the equilibrium constant for the formation reaction is called formation constant Kf [products]/[reactants]
-adding Ligands that form stable complex ions with metal cations, increase the solubility by removing free metal ions from solution
Percipatates
Q=Ksp, solution saturated no precipitation
Q<Ksp reaction goes to products solution is unsaturated no precipitation
Q>Ksp = reaction goes to reactants solution would be super saturated and precipitation
Selective precipitation
-Compound (same disassociation stoichiometry) with smaller ksp will precipitate first
