3.3.12- Acids and Bases (PAPER 1)

Question 1

What is a Bronsted- Lowry acid?

Answer 1

A proton donor (H+)

When we mix acids with water, H+ ions are released.

Question 2

Use HA to show how H+ ions are formed.

Answer 2

HA(aq) + H2O(l) —–> H3O+ (aq) + A- (aq)

Question 3

What is a Bronsted-Lowry base?

Answer 3

A proton acceptor

When we mix bases with water, they react with H+ ions to make OH- hydroxide ions.

Question 4

Use B for base to show how hydroxide ions are formed

Answer 4

B (aq) + H2O (l) ——> OH- (aq) + BH+ (aq)

Question 5

What is the Bronsted-Lowry acid-base reaction?

Answer 5

A reaction involving the transfer of a proton

Question 6

What do H+ ions form in water?

Answer 6

Hydroxonium ions H3O+, these ions make a solution acidic.

Question 7

What is the difference between strong and weak bases?

Answer 7

Strong bases are fully ionised in solution to convert into OH- ions whereas weak bases do not convert fully into OH- ions in solution.

Question 8

What is the difference between strong and weak acids?

Answer 8

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.

Question 9

What are examples of weak acids?

Answer 9

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.

Question 10

What are examples of strong acids?

Answer 10

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.

Question 11

What are examples of strong bases?

Answer 11

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.

Question 12

What are examples of weak bases?

Answer 12

NH3- ammonia

NH3 + H2O ⇌ NH4+ + OH-

Backwards reaction favoured with weak bases, low conc of OH- ions produced, equilibrium on left.

Question 13

What is generic example of acid (HA) donating a proton to a base (B)?

Answer 13

HA (aq) + B (aq) ⇌ BH+ (aq) + A- (aq)

Question 14

How does water act when an acid is added to it?

Answer 14

As a base- it accepts a proton.

HA(aq) + H2O (l) ⇌ H3O+ (aq) + A- (aq)

Question 15

How does water dissociate?

Answer 15

It dissociates very weakly into OH- and H+ ions.

Equation
2H2O ⇌ H3O+ + OH-
Simplified equation
H2O ⇌ OH- + H+

Question 16

What would kc equation for water be and why is this inaccurate?

Answer 16

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.

Question 17

What is ionic product of water?

Answer 17

Multiply the 2 previous constants- Kc and H2O to get:

Kw= [H+][OH-]
units= mol2dm-6

Question 18

What is the value of Kw at a given temperature?

Answer 18

Kw is the same in a solution at a given temperature: 1.00x10^-14 mol2dm-6

The value changes if temperature changes

Question 19

What is the conc of H+ and OH- ions in pure water?

Answer 19

They are equal
[H+]=[OH-]

This can be simplified to: Kw= [H+]^2 in pure water

Question 20

What is the equation to calculate pH?

Answer 20

pH = −log10 [H+]

Question 21

Calculate the pH of 0.03moldm-3 HCl

Answer 21

HCl dissociates fully so we assume conc of H+ = conc of HCl

pH= -log (0.03) = 1.52

Question 22

Calculate the concentration of H+ ions of nitric acid with pH 1.7

Answer 22

Inverse the pH equation to give
[H+]= 10^-pH

[H+]= 10^(-1.7) = 0.020 moldm-3

Question 23

What are monoprotic acids? Calculating pH of strong acid

Answer 23

These dissociate to produce one H+ ion for every acid molecule

Concentration of the acid= concentration of H+ ions

Eg HCl or HNO3

Question 24

What are diprotic acids? Calculating pH of strong acid

Answer 24

These dissociate to produce 2H+ ions for every acid molecule

Concentration of acid= 2x concentration of H+ ions

Eg H2SO4

Question 25

Calculate pH of 0.25moldm-3 H2SO4

Answer 25

pH= -log10 (H+) 0.25x2=0.50 -log(0.50) = 0.30

Question 26

How to calculate pH of strong base?

Answer 26

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

Question 27

Calculate pH of 0.15moldm-3 NaOH at 298K, Kw= 1x10^-14 mol2dm-6

Answer 27

Kw= [H+][OH-] [H+]= Kw/ [OH-] = 1x10^-14/ 0.15= 6.6x10^-14 pH= -log(6.6) =13.17

Question 28

How to use Ka to work out pH of weak acids?

Answer 28

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]

Question 29

Calculate the pH of 0.0300 moldm-3 CH3COOH at 298K. Ka is 1.76x10^-5moldm-3

Answer 29

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

Question 30

Calculate concentration of HCOOH at 298K with pH 3.14. Ka for methanoic acid is 1.77x10^-4moldm-3

Answer 30

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

Question 31

What is pKa used for?

Answer 31

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

Question 32

How to calculate Ka?

Answer 32

10 ^-pKa

Question 33

Calculate pH of 0.0250moldm-3 ethanoic acid at 298K with pKa value 2.75

Answer 33

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