Topic 12: Acid-Base Equilibria Flashcards
Remember that you MUST do all tasks in the acid base chemsheets booklet
What is the Bronstead-Lowry definition for acids and bases?
Acids are proton donors (H+) and bases are proton acceptors.
To accept a proton, a base must have a lone pair of e to form a dative cv bond w the proton. So a base must have an atom on the right side of the periodic table.
What are conjugate base pairs?
Use HCl + H20 –> H3O+ + Cl- in your answer
In the forward reaction, HCl is an acid bc it’s donating a proton to H20. H20 is a base bc it’s accepting a proton from HCl.
In the reverse reaction H3O+ behaves as an acid because It’s donating a proton to Cl-. Cl- is the base because it’s accepting a proton. Therefore HCl and Cl-, H3O+ and H20. Both of these are conjugate base pairs
Identify the acid and base in each of the following reactions:
State definitions for monoprotic and diprotic acids
Monoprotic acids: can only donate one proton, eg HCl
Diprotic acids: can donate 2 protons, eg H2SO4
State definitions for monoprotic and diprotic/diacidic bases
monoprotic base: accepts one proton
diprotic/diacidic base: accepts 2 protons, eg C03 2-
What is the difference between strong and weak acids?
Strong acids almost completely disassociate in aqueous solution.
Weak acids only partially dissociate in aqueous solution.
H+ conc is directly related to acid concentration. E.g. a HCl solution of 0.1moldm^3 will produce a H+ ion conc of 0.1moldm3
What is pH?
pH= -log10 [H+]
This can be rearranged to:
[H+]= 10^-pH
Calculate the pH of a solution formed when 100 cm³ of of water is added to 50 cm³ of 0.1moldm3 HN03
[H+] in original solution= 0.1
[H+] in dilute solution= 0.1 x old volume/new volume
= 0.1 x 50/150= 0.333..
pH= -log10[0.333..]
pH= 1.48
Calculate the pH of the solution formed when 250 cm³ of 0.3 moldm3 H2SO4 is made up to 1000 cm³
solution with water
[H+] in original solution equals 0.3 x 2= 0.6
You multiply by two because H2SO4 is diprotic!
[H+] in dilute solution= 0.6× 250/1000=0.15
pH= -log10 [0.15]= 0.82
What do you assume when finding the pH of strong acids?
Assume that [H+] = [HA]
How do you find [H+] of a weak acid using the acid dissociation constant?
If HA represents a weak acid the eqn for its disassociation into aq solution is: HA —> H+ + A-.
Using eqm law to find Ka: Ka= [H+(aq)] [A-(aq)] / [HA(aq)]
Assume the conc of H+ and A- is the same. This simplifies the expression to: Ka = [H+]^2 / [HA]
Therefore √Ka [HA] = H+
Calculate pH of an aqueous solution of ethanoic acid of 0.05moldm3. The value of Ka for ethanoic acid is 1.74 x 10^-5 moldm3 at 298K.
CH3COOH(aq)—-> CH3COO- + H+
What is Kw?
Pure water self ionises: H20(l) —-> H+(aq) + OH-(aq)
Applying the law of eqm: [H+] [OH-] / [H20] =Kw
As water is constant at a given temperature, the expression may be simplified to: [H+] [OH-] =Kw
298K, Kw= 1x 10^-14 mol2dm-6
What is the relationship between Kw and pKw?
pKw = -log10 Kw
Describe what happens to the pH of acids when diluted
Strong acids: PH increases by one for each tenfold decrease in concentration
Weak acids: PH increases by 0.5 for each tenfold decrease in concentration
What is the difference between the endpoints and the equivalence point of a titration?
Endpoint: When the indicator changes colour
Equivalence point: when acid + base react juntos in exact propotions.
pH at the equivalence point depends on the acid and base, e.g. strong acid +weak base will have pH<7 at the equivalence point.
What are indicators?
Most are weak acids which dissociate to form an ion of a DIFF colour to the acid molecule: Hind(aq) —> H+ + ln-(aq).
HInd and In- have diff colours in solution. For methyl orange these are red and yellow. When H+ is large eqm shifts left so solution turns red in acids. Sera a stage where [Hind]= [Ind-] and the indicator looks orange. Find the exact pH of this using the eqm constant for methyl orange:
KInd= [H+] [in-] / [Hind] = 2 x 10^-4 moldm3
When HInd= Ind-, expression simplifies: Kind= [H+]= 2 x 10^-4. This gives a pH 3.7 at the halfway stage and is the same value as pKind
The best indicator to choose for a titration has a pKind value closest to the pH at the equivalence point
Describe and explain the titration curve for:
Strong base + Strong acid
Any indicators within the straight line can be used in the titration. this graph has a longer straight line
Describe and explain the titration curve for:
Titration of a strong acid with a weak base
The straight line happens at neutralisation. It doesn’t reach such a high pH bc it is a weak base. Straight line is shorter.
Titration curve to show the titration of a weak acid with a strong base?
The dashed line is the same as the end point.
What does a titration curve of a weak acid with a weak base look like?
There is no steep section in this graph. Instead hay a point of inflection. The lack of a steep section means that its difficult to do a titration using an indicator.
Calculate the pH of 0.1moldm3 H2SO4
2 x 0.1= 0.2 (DIPROTIC!)
-log10 [0.2]= 0.698 = 0.70
Calculate the pH of 0.25moldm-3 methanoic acid. Ka= 1.7 x 10^-4
[H+]= Square root of Ka[HA]
= Square root of (1.7 x 10^-4 x 0.25)= 6.5 x 10^-3
pH= -log10 [6.5 x 10^-3]
= 2.19
What are buffers?
A solution which resists any change in pH when a small amount of acid or base is added, so pH remains almost constant. Hay acid and alkalai buffer solutions:
Acid buffer solutions: mixture of a weak acid and its conjugate base of similar conc, eg enthanoic acid and Na ethanoate.
Alkaline buffer solutions: a mixture of a weak base and its conjugate acid of similar conc, eg ammonia and ammonium chloride.
