Acid-Base And pH (w4) Flashcards
(43 cards)
pKa shows …
How strong an acid is
Strong acid =
Completely ionised in water
pH =
-log10[H+]
Strong base =
Completely dissociated
Kw = 1 x 10^(-14)
Kw =
[H+][OH-]
Weak acids =
Don’t completely ionise in water.
The amount of ionisation is governed by the equilibrium.
pKa?
The pH at which it is exactly half dissociated.
If pH is increased (more basic) ?
The acid will become more ionised, pH > pKa dissociated.
If pH is lowered (more acidic) ?
The acid becomes less ionised, pH < pKa undissociated.
Ratio of ionised: non-ionised forms …
A shift in pH by one unit to either side of the pKa value must change the ration by a factor of 10.
Outline an equation for a neutralisation reaction …
ACID + BASE —> SALT + WATER
What does salt hydrolysis usually effect ?
The pH of a solution
Strong acid + strong base =
Neutral salt
PH = 7
Strong acid + weak base =
Acidic salt
pH <7
Weak acid + strong base =
Basic salt
pH >7
Weak acid + Weak base =
Dependant on which one is stronger
Weak acid + weak base concepts :
- Ka cation > Kb anion = acidic solution
- Kb anion > Ka cation = basic solution
- if Ka and Kb are similar the solution is close to neutral.
What is the half equivalence point ?
The half point - the half of the total amount of base (added) needed to neutralise the acid it has been added to.
PH = pKa of the weak acid
What us the equivalence point ?
The end point of the reaction = there are equal stoichiometric amounts.
The reaction has completed as all the acid has been converted to its conjugate Base.
What is an acid-base titration (neutralisation) ?
The quantitative analysis of the concentration of an unknown acid or base solution.
What us a buffer solution ?
A solution which is able to resist changes in pH on addition of small amounts of acid or base.
Define buffer capacity ….
The number of moles per litre of a strong acid or base required to produce an increase or decrease of one pH unit in the solution.
IN CLASS TEST ;
What is the pH of a buffer solution containing 0.042M NaH2PO4 and 0.058M Na2HPO4 ?
HINT
Use the Henderson-Hasselbalch equitation;
PH = pKa + log10 x ([A-]/[HA-])
Figure out which one is the base [A-]
And which is the acid [HA-]
What do we do to move the pH of the buffer solution up by one unit of pH = pKa ? (More basic)
We need to insure the ionised form is 10 times higher.