3.3.12- Acids and Bases (PAPER 1) Flashcards
What is a Bronsted- Lowry acid?
A proton donor (H+)
When we mix acids with water, H+ ions are released.
Use HA to show how H+ ions are formed.
HA(aq) + H2O(l) —–> H3O+ (aq) + A- (aq)
What is a Bronsted-Lowry base?
A proton acceptor
When we mix bases with water, they react with H+ ions to make OH- hydroxide ions.
Use B for base to show how hydroxide ions are formed
B (aq) + H2O (l) ——> OH- (aq) + BH+ (aq)
What is the Bronsted-Lowry acid-base reaction?
A reaction involving the transfer of a proton
What do H+ ions form in water?
Hydroxonium ions H3O+, these ions make a solution acidic.
What is the difference between strong and weak bases?
Strong bases are fully ionised in solution to convert into OH- ions whereas weak bases do not convert fully into OH- ions in solution.
What is the difference between strong and weak acids?
Strong acids dissociate/ ionise completely in aqueous solutions to produce high conc of H+ ions whereas weak acids only partially dissociate to produce low conc of H+ ions.
What are examples of weak acids?
Carboxylic acids such as CH3COOH (ethanoic acid)
CH3OOH ⇌ CH3COO- + H+
Backwards reaction is favoured with weak acids so not many H+ ions produced, equilibrium far on the left.
What are examples of strong acids?
HCl- hydrochloric acid
H2SO4- sulfuric acid
HNO3- nitric acid
HCl ⇌ H+ + Cl-
Forwards reaction is favoured with strong acids, dissociate almost completely into a large conc of H+ ions, equilibrium on right.
What are examples of strong bases?
NaOH- sodium hydroxide
KOH- potassium hydroxide
NaOH ⇌ Na+ + Cl-
Forwards reaction is favoured with strong bases, large conc of OH- ions produced, equilibrium on right.
What are examples of weak bases?
NH3- ammonia
NH3 + H2O ⇌ NH4+ + OH-
Backwards reaction favoured with weak bases, low conc of OH- ions produced, equilibrium on left.
What is generic example of acid (HA) donating a proton to a base (B)?
HA (aq) + B (aq) ⇌ BH+ (aq) + A- (aq)
How does water act when an acid is added to it?
As a base- it accepts a proton.
HA(aq) + H2O (l) ⇌ H3O+ (aq) + A- (aq)
How does water dissociate?
It dissociates very weakly into OH- and H+ ions.
Equation
2H2O ⇌ H3O+ + OH-
Simplified equation
H2O ⇌ OH- + H+
What would kc equation for water be and why is this inaccurate?
Kc= [H+] [OH-]/ [H2O]
The conc of H+/OH- is so little compared to H2O we can assume that conc of H2O has constant value.
What is ionic product of water?
Multiply the 2 previous constants- Kc and H2O to get:
Kw= [H+][OH-]
units= mol2dm-6
What is the value of Kw at a given temperature?
Kw is the same in a solution at a given temperature: 1.00x10^-14 mol2dm-6
The value changes if temperature changes
What is the conc of H+ and OH- ions in pure water?
They are equal
[H+]=[OH-]
This can be simplified to: Kw= [H+]^2 in pure water
What is the equation to calculate pH?
pH = −log10 [H+]
Calculate the pH of 0.03moldm-3 HCl
HCl dissociates fully so we assume conc of H+ = conc of HCl
pH= -log (0.03) = 1.52
Calculate the concentration of H+ ions of nitric acid with pH 1.7
Inverse the pH equation to give
[H+]= 10^-pH
[H+]= 10^(-1.7) = 0.020 moldm-3
What are monoprotic acids? Calculating pH of strong acid
These dissociate to produce one H+ ion for every acid molecule
Concentration of the acid= concentration of H+ ions
Eg HCl or HNO3
What are diprotic acids? Calculating pH of strong acid
These dissociate to produce 2H+ ions for every acid molecule
Concentration of acid= 2x concentration of H+ ions
Eg H2SO4
Calculate pH of 0.25moldm-3 H2SO4
pH= -log10 (H+)
0.25x2=0.50
-log(0.50) = 0.30
How to calculate pH of strong base?
Most dissociate to produce 1 OH- ion for each base molecule
Concentration of the base= concentration of OH- ions
Use Kw= [H+][OH-]
Have to know Kw and [OH-] at given temperature to find [H+] then pH
Calculate pH of 0.15moldm-3 NaOH at 298K, Kw= 1x10^-14 mol2dm-6
Kw= [H+][OH-]
[H+]= Kw/ [OH-]
= 1x10^-14/ 0.15= 6.6x10^-14
pH= -log(6.6) =13.17
How to use Ka to work out pH of weak acids?
Cannot assume that [H+]= [acid]
Small amount of weak acid (HA) dissociates so can assume that [HA] at equilibrium= [HA] at the start
Ka(moldm-3) = [H+] [A-]/ [HA] at start
Dissociation of acid is greater than dissociation of water present in solution- all H+ ions come from acid
[H+] = [A-]
Ka= [H+]^2/ [HA]
Calculate the pH of 0.0300 moldm-3 CH3COOH at 298K. Ka is 1.76x10^-5moldm-3
Write down Ka expression
Ka= [H+]^2 / CH3COOH
Rearrange for [H+]^2
[H+]^2 = Ka x [CH3COOH]
= 5.28x10^-7
[H+]= square root of 5.28^-7= 7.27x10^-4moldm-3
pH= -log(7.27x10^-4)= 3.14
Calculate concentration of HCOOH at 298K with pH 3.14. Ka for methanoic acid is 1.77x10^-4moldm-3
Calculate [H+] by using pH= -log [H+]
10^-pH = 10^-3.14= 7.24x10^-4
Ka= [H+]^2/ HCOOH
Rearrange for HCOOH
HCOOH= [H+]^2/ Ka
=2.96x10^-3moldm-3
What is pKa used for?
Another way of measuring the strength of an acid, lower value= stronger acid.
pKa= -log10 (Ka)
Eg pKa of acid with Ka 7.52x10-3=
-log(7.52x10-3) = 2.12
How to calculate Ka?
10 ^-pKa
Calculate pH of 0.0250moldm-3 ethanoic acid at 298K with pKa value 2.75
Calculate Ka
Ka= 10^-pKa = 10^-2.75= 1.78x10^-5moldm-3
Calculate [H+] from Ka expression
[H+]^2/ [CH3COOH]
[H+]^2= Ka x [CH3COOH]
=1.78x10^-5 x 0.0250= 4.45x10^-7
Square root answer= 6.67x10^-4
pH= -log10[H+]
= 3.18
