Acids and bases

What is a bronstead Lowry acid

a proton donor

H₃O⁺

hydronium ion

What is a monoprotic and dirpotic acid?

release one mole of H+ per mole of acid

nHA = nH+
Release two moles of H+ per mole of acid

2nH₂A = nH+

Define weak and strong acid

one that partially dissociated into ions in aqueous solutions

one that fully dissociates into ions in aqueous solutions 

Write dissociation equation of methanol acid

HCOOH →← HCOO- + H+

What does the reversible reaction arrow show?

denotes partial dissociation

equilibrium lies very much to the left so there is a low conc of protons

What is a bronstead Lowry base?

proton acceptor

examples of bases

metal hydroxides/oxides/carbonates

ammonia

amines

Equation and ph of metal oxides with water Na₂O

Na₂O + H₂O → 2NaOH

pH 14

Equation and ph of ammonia

NH₃ + H₂O →← NH₄⁺ + OH^-

pH 9

What is a mono basic base?

release of one mole of OH- ions per mole of base which reacts with one mole of H+ ions

Q: Explain why the pH formed with CH₃NH₂ is added to H₂O is not higher (2)

equilibrium lies to the left so low concentration of OH-

Define strong bases and examples

Define weak bases

proton acceptors that fully dissociate into ions in aqueous solution

sodium hydroxide

proton acceptors that partially dissociate into ions in aqueous solution

ammonia and amines

Acid base equilbira equation 

HA + B →← BH+ + A-

acid + base → conjugate acid + conjugate base 

How does a conjugate base/acid form?

when an acid donates a proton/when a base accepts a proton

Define conjugate pair

a pair of species related by the loss or gain of a proton 

name acid and conjugate acid in these reactions:

1. NH₃ + H₂O →← NH₄⁺ + OH^-

2. CH₃COOH + H₂O →← CH₃COO^- + H₃O⁺

3. H₂SO₄ + HNO₃ →← HSO₄^- + H₂NO₃⁺

1. water , NH4+

2. ethanoic acid, H₃O⁺

3. sulfuric acid as it is stronger, H₂NO₃⁺

conjugate pair practice pg8

s

Define/state what is meant by pH

pH = -log[H⁺]

1. calculate pH of [H+] =0.125moldm^-3

2. calculate [H+] of pH=2.50

0.90

3.16×10^-3

Calculate pH of 0.005moldm^-3 of [sulfuric acid]

0.005 × 2 (diprotic)

pH = 0.70

pH calc practice pg12

s

Dilution calc pg 13-14 + paper

Calculate the pH of the solution formed when 250cm³ of 0.300moldm^-3 H₂SO₄ is made up to 1000cm³ solution with water

n sulfuric acid

diprotic so x2

new conc of protons

pH = 0.82

What volume of water would need to be added to 35cm³ of 0.185moldm^-3 HCl to form a solution of pH 1.29?

conc of protons

moles of HCl = mol of H+

calc volume

volume -35 = 91cm³

Calculate the pH of the solution formed when 60cm³ of water is added to 90cm³ of 0.195moldm^-3 LiOH (Kw = 1×10^-14

n LiOH 

[nLiOH] = [nOH^-]

new [OH-] 

use Kw

pH = 13.07

What volume of water must be added to 160cm³ of a 0.395moldm^-3 aqueous solution potassium to make a solution of pH 12.71. Kw = 1.00×10^-14 mol²dm^-6

1071cm³

extra paper questions

Equation for dissociation of water

H₂O →← H⁺ + OH^-

Q: why is H₂O not in the K_c expression for the dissociation of H₂O?

H₂O →← H⁺ + OH^-

equilbirum lies to the left

as conc of water is very much greater than the conc of protons and OH-, [H₂O] is effectively a constant number

What is Kw?

ionic product of water 

=[H+][OH-]

Q:Why is dissociation of water endothermic?

it involves the breaking of bonds

Q: Explain how increasing temperature affects value of Kw?

equilbirum shifts right as the forward reaction is endothermic to oppose the increase in temperature. More H+ and OH- will be produced and value of Kw increases so pH of water decreases at higher temperature

Kw calc pg16-18

Calculate the pH of water when Kw = 5.13×10^-14

pH = 6.64

Calculate the pH of 0.1moldm^-3 solution of calcium hydroxide 

pH = 13.30

(0.1 × 2)

Calculate the concentration of an aqueous solution of Ba(OH)₂ with a pH of 13.30 at 298K

find [OH-] using Kw

conc = 0.0998moldm^-3

^{(divide by 2)}

Strong acid and strong base reaction method

practice on pg20

find which is in excess

Calculate the pH of the solution formed when 50cm3 of 0.100moldm-3 H2SO4 is added to 25cm3 of 0.150moldm-3 NaOH

n of NaOH = nOH^-

nH2SO4 ×2 = nH⁺

nH⁺ is in XS - find moles

find conc of H⁺

pH = 1.08

what is Ka?

the acid dissociation constant 

Write an equation for Ka of

1. H₂CO₃

2. NH₄⁺

3. H₂S

= [H⁺] [HCO₃^-] / [H₂CO₃]

= [H⁺][NH₃^-] / [NH₄⁺]

= [H⁺][HS^-] / [H₂S]

Q: Ethanoic acid is a weak acid. At 298K, Ka=1.74×10^-5 moldm^-3
Explain what happens to the value of Ka when the temperature is increased (3)

Dissociation is endothermic

As temperature increases the value of Ka increases as equilbrium will shift to the right to oppose the increase in temperature.

Concentration of H⁺ ions will increase and the pH will decrease

Q: Why do strong acids not have a value for Ka?

they fully dissociate

Ka calc pg24-25

A 0.100moldm-3 solution of a weak acid HA has a pH value of 2.50. Calculate Ka

1×10^-4moldm^-3

How to find pKa (using Ka) and Ka (using pKa)?

pKa = -logKa

Ka = 10^-pKa

Calculating pH of solution formed from weak acids and strong base technique

Pratcice 27-28

if XS of OH- then use Kw

is XS acid use Ka

Use RICE 

pg27 

Calculate pH of the solution when 30cm3 of 0.200moldm-3 ethanoic acid (pKa = 4.76) is added to 100cm3 of 0.100moldm-3 NaOH 

USE RICE

XS NaOH → Kw

pH = 12.49

Calculate pH of the solution when 50cm3 of 0.500moldm-3 ethanoic acid (pKa = 4.76) is added to 75cm3 of 0.200moldm-3 NaOH 

use RICE

XS acid → find Ka

pH = 4.94

( n of CH₃COO^- is in the RICE table - not 0)

Calculate pH of the solution when 100cm3 of 0.200moldm-3 ethanoic acid (pKa = 4.76) is added to 40cm3 of 0.250moldm-3 KOH and state what happens

RICE TABLE

XS acid → Ka

[H+] = Ka

Half neutralisation

pH = pKa

pH = 4.76