A summary of topic 12, organised so the notes are easy to understand. The notes are on slides, so they can be printed out and used as revision cards or posters, for revision on the go. The notes cross-reference the specification so it is easy to see where each bit of information has come from. They...
Brønsted–Lowry acid–base theory: pH of weak acids:
- Acids = proton (H+) donors = partially dissociate into ions in solution
- Bases = proton (H+) acceptors CH3COOH(aq)⇌ CH3COO-(aq) + H+(aq)
Conjugate acid-base pairs = 2 species that Assumptions:
differ by H+ - [CH3COOH]i = [CH3COOH]eqm
E.g HA + H2O ⇌ A- + H3O+ - [CH3COO-]=[H+]
- Pair 1 = HA ⇌ A- Equations:
- Pair 2 = H2O ⇌ H3O+ - Ka = [CH3COO-][H+] = [H+]^2
pH of strong acids: [CH3COOH] [CH3COOH]
= completely dissociate into ions in - pH = -log10[H+]
solution At half neutralization point, pKa = pH
HA(aq) → H+(aq) + A-(aq) pH of buffers:
= Strong acids start flat
When pH is neutral [H+] = [OH-] Equations:
= Weak acids start with a flick
Assumptions: - Ka = [CH3COO-][H+]
Finding out where pH rise is:
- [HA]=[H+] - full dissociation [CH3COOH]
E.g. 10cm^3 of 0.2moldm^-3 HNO3
- Or [H2A]=[H+]/2 - pH = -log10[H+]
and 50cm^3 of 0.05moldm^-3 NaOH
Equations: - Find conc of each thing using Ci x Vi
Base is being added ∴ find volume
- pH = -log10[H+] Vtot
of base needed to neutralize acid
pH of strong bases: Assumptions:
HNO3 + NaOH
=completely dissociate into ions in solution - [CH3COONa]=[CH3COO-]
C= 0.2 0.05
BOH(aq) →B+(aq) + OH-(aq) - [CH3COOH] = conc of acid
X4
When pH is neutral [H+] = [OH-] pKa and pKw:
∴v(HNO3) x4 = v(NaOH) needed
Assumptions: pKa = shows how weak or strong an acid
= 10x4=40cm^3
- [MOH]=[OH-] is, lower pKa = stronger acid
Choosing an indicator:
Equations: pKw = the negative logarithm of the ionic
Changes colour within the pH range
- Kw = [H+][OH-] Kw=1x10^-14 product of water (Kw)
of the vertical section of the curve.
- pH = -log10[H+] Titration curves:
Hard to find for weak acids/bases -
pH of water: Equivalence point = when you get a
no vertical section.
Assumptions: neutral solution [OH-]=[H+]
- [H+] = [OH-] End point = indicator colour change
Equations: Titrations assume end point = equivalence
- Kw = [H+][OH-] point
- pH = -log10[H+]
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