IONIC EQUILIBRIA Flashcards
Kw
‣ ionic product of water
H₂O ⇌ H⁺ + OH⁻
Kc=[H⁺][OH⁻]/[H₂O]
Kw=[H₂O]*Kc
Kw = [H⁺] [OH⁻]
monoprotic/monobasic acids
acids which release one proton and is neutralised by 1 mol of a base
‣ HCl → H⁺ + Cl⁻
‣ HNO₃ → H⁺ + NO₃⁻
‣ CH₃COOH ⇌ H⁺ + CH₃COO⁻
diprotic/dibasic acids
acids, when upon complete neutralisation, give out 2 protons
H₂SO₄ → 2H⁺ + SO₄²⁻
neutral salt
HCl + NaOH → NaCl ₊ H₂O
complete neutralisation
acidic salt
‣ H₂SO₄ + NaOH → NaHSO₄ (partial neutralisation, acidic salt so reacts again with 1 more mole of base)
‣ NaHSO₄ + NaOH → Na₂SO₄ + H₂O
how to find pH of strong acids
pH = -log₁₀[H⁺]
‣ H₂SO₄ → 2H⁺ + SO₄²⁻
0.01, 0.02, 0.1
‣ pH=-log₁₀[0.02]=1.7
how to find pH of strong bases (using Kw)
NaOH → Na⁺ + OH⁻
0.05, 0.05, 0.05
Kw=[H⁺][OH⁻]
H⁺ = Kw/[OH⁻]
= 110⁻¹⁴/0.05 =210⁻¹³
pH = -log₁₀[H⁺]
= -log₁₀[2*10⁻¹³]
= 12.7
how to find pH of strong bases (using pOH)
NaOH → Na⁺ + OH⁻
0.05, 0.05, 0.05
pOH = -log₁₀[OH⁻]
= -log₁₀[0.05]=1.3
pH + pOH = 14
pH = 14-1.3 = 12.7
how to find pH of weak acids
CH₃COOH ⇌ H⁺ + CH₃COO⁻
·Kc=[H⁺][CH₃COO⁻]/[CH₃COOH]
·Kc→Ka (acid dissociation constant)
·Ka=[H⁺]²/[CH₃COOH]
H⁺= √Ka * [CH₃COOH]
pH = -log₁₀[H⁺]
if Ka is higher
it’s a stronger acid (faster forward rate, more ionisation)
effect of electron releasing groups on a weak acid (+I)
· the presence of CH₃ (alkyl/electron-releasing groups) creates a positive inductive effect
· it increases the negative charge making the (ethanoate) ion unstable
· the equilibrium shifts to LHS; less ionisation
· the more alkyl groups attached, the weaker the acid
· HCOOH ⇌ H⁺ + HCOO⁻ (no electron releasing groups so stronger)
· CH₃COOH ⇌ H⁺ + CH₃COO⁻ (has an alkyl group which releases charge, causing the ion to become unstable and reacts with H⁺ again and forms CH₃COOH)
effect of electron withdrawing groups on the acid (-I)
- presence of a more electronegative atom such as Cl increases the acidity due to it pulling the electron pair towards it
- the resulting unequal sharing of electrons generates a positive charge which is transmitted through the chain
- Cl atoms are electron withdrawing so they reduce the negative charge on the propanoate ion hence increases the stability, more ionisation
- negative charged molecules are always less stable
strong acid + strong base titration end-point
pH = 7
strong acid + weak base titration end-point
pH = below 7 (indicator in acidic range)
weak acid + weak base titration end-point
pH = 7
weak acid + strong base titration end-point
pH = above 7 (alkaline range)
definition of a buffer
substance that resists changes in pH (when small amounts of an acid or alkali are added)
what is a buffer made of?
- mixture of weak acid & its conjugate base
buffer made out of (CH₃COOH ⇌ H⁺ + CH₃COO⁻) would be CH₃COOH/CH₃COO⁻ (needs to be of equal/comparable concentration)
or
- mixture of weak base & its conjugate acid
buffer made out of (NH₃ + H⁺ ⇌ NH₄⁺) would be NH₃/NH₄⁺
how to make a buffer solution of CH₃COOH/CH₃COO⁻
· weak acids/bases are partially ionised so its conjugate base/acid does not match in concentration
· mix the weak acid (CH₃COOH/1.5 moldm⁻³) and a salt of ethanoic acid - the salt provides the ethanoate ion (conj. base in exact amount)
CH₃COONa → CH₃COO⁻ + Na⁺
∴ buffer consists of weak acid + salt of acid
uses of buffers
- shampoos, creams
- nappy rash creams
- treatment of leather
- making dyes
calculate pH of a buffer
Kₐ = [H⁺][CH₃COO⁻]/[CH₃COOH]
H⁺ = Kₐ [CH₃COOH]/[CH₃COO⁻] take log on all sides
-log₁₀[H⁺] = -log₁₀Kₐ - log₁₀[CH₃COOH]/[CH₃COO⁻]
pKₐ = -log₁₀Kₐ
pH = pKₐ + log₁₀ [salt/acid]
or
pH = pKₐ - log₁₀ [acid/salt]
high pKₐ
weak acid (low Kₐ)
high Kₐ
strong acid (low pKₐ)
Ksp (solubility product)
AgCl ⇌ Ag⁺ + Cl⁻ (“insoluble”)
!!! rate at dissolving is equal to the rate of precipitation
Kc = [Ag⁺][Cl⁻]/[AgCl]
K * [AgCl] = Ksp
Ksp = [Ag⁺][Cl⁻]
