JF Basic Chemistry Tutorial : Acids

1
JF Basic Chemistry Tutorial Acids Bases

• Shane Plunkett
• plunkes_at_tcd.ie

• Acids and Bases
• Three Theories
• pH and pOH
• Titrations and Buffers

• Recommended reading
• M.S. Silberberg, Chemistry, The Molecular Nature
  of Matter and Change

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Acids and Bases

• Dealing with ions
• Arrhenius theory an acid which contains
  hydrogen and can dissociate in water to produce
  positive hydrogen ions
• e.g. HX H2O ? H (H3O) X-
• A base reacts with a protonic acid to give water
  (and a salt)
• e.g. HCl NaOH ? NaCl H2O
• BrØnsted-Lowry Theory acids are proton donors
  bases are proton acceptors
• e.g. HCN H2O ? H3O CN-
• HCN is an acid, in that it donates a proton to
  water. Water is acting as a base, as it accepts
  that proton
• Lewis Theory an acid accepts a pair of
  electrons a base donates a pair of electrons
• e.g. HCl NaOH ? NaCl H2O
• The strength of an acid depends on the extent to
  which is dissociates and is measured by its
  dissociation constant

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Strong and weak acids and bases

• Strong acid fully dissociates in water, i.e.
  almost every molecule breaks up to form H ions
• Some strong acids areHCl, H2SO4, HNO3
• Weak acid partially dissociates in water
• Some weak acids arecarboxylic acids such as
  CH3COOH, C2H5COOH
• Strong base fully dissociates in water, i.e.
  almost every molecule breaks up to form OH- ions
• Some strong bases are.NaOH, compounds which
  contain OH- ions or O2- ions
• Weak base partially dissociates in water
• Some weak basesnitrogen-containing compounds,
  such as NH3
• Strengths can be determined by the acid or base
  dissociation constant

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Acids

• Act as proton donors
• Electron pair acceptors
• Strong acids dissociate fully in water.
• Weak acids partially dissociate.
• Ka acid dissociation constant
• HA H2O ? H3O A-
• Ka H3OA-
• HA
• Higher Ka values mean stronger acids

Bases

• Act as proton acceptors
• Electron pair donors
• Strong bases dissociate fully in water
• Weak bases partially dissociate
• Kb base dissociation constant

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pH and pOH

• H3O can vary greatly ? logarithmic scale used
• pH -log H3O
• pOH -log OH-
• pH gt 7 basic
• pH 7 neutral
• pH lt 7 acidic
• Can also express dissociation constants in terms
  of logs pKa -log Ka
• ? the higher the Ka the lower the pKa
• Similarly for bases

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Ionic product of water
2H2O ? H3O OH-
Kw is the ion-product constant. Kw is the
product of the molar concentrations of H3O and
OH- ions at a particular temperature
Kw H3OOH-
________ H2O2
H2O is constant
Kw H3OOH- 1.0 10-14 at 25C
Solution is neutral
H3O OH-
H3O gt OH-
Solution is acidic
pKw -log Kw
H3O lt OH-
Solution is basic

• Can incorporate pH and pOH

-log Kw -log H3OOH-
Kw H3OOH-
pKw pH pOH 14 (at 25C)
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Titrations

• Acid-base stoichiometric point can be determined
  by using an indicator
• Indicators are dyes whose colour depends on the
  pH
• HIn is protonated form and In- is its conjugate
  base
• HIn H2O ? H3O In-
• KIn H3OIn-
• HIn
• If pH goes from low to high, the indicator goes
  from HIn to In- and a colour change is noted

Strong acid-base curve
Strong base-weak acid
Strong acid-weak base
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Buffers

• Allow pH to be maintained over small additions of
  acid or base
• Made up of a weak acid and its conjugate base,
  e.g.
• HA H2O ? A- H3O
• acid conjugate base
• The equilibrium will shift to the right on
  addition of a small amount to base and shifts to
  the left on addition of small amounts of acid
• Henderson-Hasselbalch equation allows
  determination of pH in buffer systems

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Questions

• What is the pH of a 0.14M aqueous solution of HCN
  (pKa 9.31)?

pKa -log Ka
Ka 4.9 10-10
HCN H2O ? H3O CN-
Ka H3OCN- HCN
HCN H3O CN-
Initial conc 0.14 0 0
Equil. Conc 0.14 x x x
x2 . 0.14-x
Ka xx 0.14-x
4.9 10-10
4.9 10-10
x2 4.9 10-10 (0.14 x)
x2 4.910-10x 6.8610-11 0
x2 6.86 10-11 4.9 10-10x
Solve for x
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x 8.2510-6 ? H3O 8.2510-6

• pH -log H3O pH -log (8.2510-6) pH 5.08

Question
The pH of 0.1M CH3COOH is 2.87. What is the
value of the acid dissociation constant, Ka?
pH -log H3O 2.87 ? H3O 1.3510-3
CH3COOH H2O ? H3O CH3COO-
Ka H3OCH3COO- CH3COOH
CH3COOH H3O CH3COO-
Initial conc 0.1 0 0
Equil. Conc 0.1 x x x
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Ka 1.3510-31.3510-3 0.1
Ka 1.8 10-5
Question
Estimate the pH of 10-7M HCl(aq)
pH -log H3O
Two contributions to H3O concentration HCl and
H2O (aqueous soln)
pH of pure water 7 pH -log H3O If -log
H3O 7 ? H3O 1 10-7
HCl H2O ? H3O Cl-
HCl is a strong acid. ? dissociates completely
in water
? H3O 10-7
Total H3O 1.0 10-7 10-7 2.0
10-7
pH -log (2.0 10-7) pH 6.7
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Calculate the concentration of OH- for an aqueous
solution with a pH of 9.60
Question

• Two contributions to OH- concentration.
• pH of pure water 7.0 ? OH- 1.0 10-7
• pH of solution 9.60
• pH pOH 14 ? pOH 14 9.60 4.4
• pOH -log OH- 4.4
• OH- 3.98 10-5
• Total OH- 3.99 10-5 ? 4.0 10-5

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2001 Paper 1 Question 8 (b)

• A buffer solution is made by adding 0.3 mol of
  acetic acid and 0.3 mol of
• sodium acetate to enough water to make 1L of
  solution. Determine the
• pH of the buffer solution. The Ka for acetic
  acid is 1.8 10-5

A- sodium acetate HA acetic acid
pH pKa log A- / HA
pH -log (1.8 10-5) log (0.3) / (0.3) pH
4.74
Calculate the pH of the buffer solution after
0.02 moles of sodium hydroxide is added
1 litre of buffer contains 0.3 moles of sodium
acetate and 0.3 moles of acetic acid. Sodium
hydroxide is a strong base. The acetic acid
will react with the base added to try to maintain
the pH. What the acid loses in concentration,
the salt (sodium acetate) will gain.
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CH3COOH OH- ? CH3COO- H2O
CH3COOH OH- CH3COO-
Initial conc 0.3 0 0.3
Change 0.3 0.02 0.02 0.3 0.02
Equil. Conc. 0.28 0.02 0.32
pH pKa log A- / HA pH - log (1.8
10-5) log (0.32) / (0.28) pH 4.74 0.058 pH
4.8