module 5 - acids, bases and buffers Flashcards
bronsted-lowry acid
a proton donor
bronsted-lowry base
a proton acceptor
strong and weak acids
fully v partially dissociate
-strong acids exist manly as ions and weak acids mainly as molecules
conjugate-acid base pairs
- contains two species that can be interconverted by transfer of a proton
- they differ by H+
hydronium ion
- dissociation doesn’t take place unless water is present
- HCl + H2O –> H3O+ + Cl-
- the hydronium ion is a conjugate acid
monobasic, dibasic…
refers to the total number of hydrogen ions that can be replaced per molecule in an acid-base reaction
redox equation for acid and a metal
2H+ + Zn –> Zn2+ + H2
ionic equation for acids and carbonates
2H+ + CO3-2 –> H2O + CO2
ionic equation for acid and metal oxides
2H+ + MgO –> Mg2+ + H2O
PH of strong acids
[H+] = [HA]
-therefore can be calculated directly
Ph of diluted strong acids
- find the PH before and after dilution
- look at change in PH
the acid dissociation constant (Ka)
Ka = [H+] [A-] / [HA]
- units: moldm-3
- changes with temperature
- the larger the Ka value, the further the equilibrium is to the right, the stronger the acid
relationship between pKa and Ka
pKa = -log(Ka)
Ka = 10 -pKa
-pKa values are more manageable than Ka values, making it easier to compare acid strengths
rules about Ka and pKa
- the stronger the acid, the larger the Ka value and the smaller the pKa value
- the weaker the acid, the smaller the Ka value and the larger the pKa value
approximations for PH of weak acids
- HA dissociates to produce equal equilibrium concentrations of H+ and A-. The dissociation of water is negligible
- [HA eqm] = [HA start] as the dissociation of weak acids is small
when do these approximations break down?
- if there is a very dilute solution (PH > 6)
- if you have a stronger weak acid as H+is too big to be considered negligible
- therefore it is true for weak acids with small Ka values
Kw
the ionic product of water (the ions in water multiplied together)
- it varies with temperature
- it has a value of 1x10-14 at 25 degrees
prove water has a PH of 7
- On dissociation, water is neutral, it produces the same number of OH- and H+ ions
- Kw = [H+][OH-]
- [H+] = square root of 1x10-14 = 1x10-7
- PH = -log[H+] = 7
strong bases
- they fully dissociate so the [OH-] = [NaOH]
- rearrange Kw = [H+][OH-]
example of weak base
- a weak base is also an alkali
- ammonia gas is an example of a weak base
- an equilibrium is set up positioned well the left.
buffer solution
a system that minimises PH changes when small amounts of an acid or base are added
-they contain a weak acid to remove added alkali and a conjugate base to remove added acids
preparing a buffer from a weak acid and its salt
- mix a weak acid with a solution of one of its alts
- when added to water, ethanoic acid partially dissociates, producing a small amount of ethanoate ions. This is the buffers solution source of weak acid.
- salts are a source of conjugate base. when added to water, it completely dissociates
- we need to add the salt as the weak acid only partially dissociates so there isn’t much conjugate base
- this creates large reservoirs of weak acid and conjugate base
preparation by partial neutralisation
-the weak acid is partially neutralised by the alkali forming the conjugate base
some of the weak acid is left over which remain unreacted
how does a buffer neutralise PH change when acid is added?
- on addition of an acid [H+] increases
- H+ ion react with the conjugate base A-
- the equilibrium position shifts to the left, removing most of the H+ ions and resulting in minimal PH change
how does a buffer neutralise the PH change when alkali is added?
- on addition of alkali, [OH-] increases
- the small concentration of H+ ions react with the OH- ions to form water
- HA dissociates, shifting the equilibrium position to the right to restore most of the H+ ion, minimising change in PH.
when are buffers most effective?
when there are equal concentrations of the weak acid and conjugate base.
what holds true when [HA] = [A-] in a buffer?
- the PH of the buffer solute is the same as the pKA value of HA
- the operating PH is typically over about two PH units, centred at the PH of the pKa value
assumptions about calculating PH from buffers
- CH3COOH has not fully dissociated [HA]start = [HA]equilibrium
- CH3COOH has fully dissociated [CH3COONa] = [CH3COO-]
how do you calculate PH of a buffer solution?
[H+] = Ka x [HA]/[A-]
why does blood act as a buffer solution?
- blood PH needs to be maintained at a PH between 7.35 and 7.45
- the carbonic acid-hydrogen carbonate (H2CO3/ HCO3-) is the most important
- if the PH slips out of this range, people can develop acidosis which can be fatal. If it goes above this PH range, people can develop alkalosis
- the b body produces more acidic material than alkaline which the conjugate base converts to H2CO3.
- The body prevents it from building up by converting H2CO3 in to carbon dioxide gas which is exhaled by the lungs.
equations for blood buffer
On addition of an acid
H2CO3 H+ + HCO3-
On addition of an alkali
H2CO3 H+ and HCO3-
