Acids, Bases and Buffers

Section 1 of 18Bronsted–Lowry acids and bases

Bronsted acid

Bronsted acid a species that can donate a proton

• HCl(aq) → H⁺(aq) + Cl⁻(aq).

Bronsted base

Bronsted base a species that can accept a proton

• OH⁻(aq) + H⁺(aq) → H₂O(l).

Conjugate acid–base pair

Conjugate acid–base pair two species that differ by one H⁺

• In CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺, CH₃COOH (acid) pairs with CH₃COO⁻ (conjugate base).
• H₂O (base) pairs with H₃O⁺ (conjugate acid).

Weak acid

• Weak acid it partially dissociates, so CH₃COOH, H₂O, CH₃COO⁻ and H₃O⁺ are all present at equilibrium

Section 2 of 18Monobasic, dibasic and tribasic acids

Monobasic (monoprotic)

• Monobasic (monoprotic) one ionisable hydrogen per molecule, e.g. HCl, HNO₃, HCN
• HCl + NaOH → NaCl + H₂O.

Dibasic (diprotic)

• Dibasic (diprotic) two ionisable hydrogens, ionising in two steps, e.g. H₂SO₄
• H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.

Tribasic (triprotic)

• Tribasic (triprotic) three ionisable hydrogens, ionising in three steps, e.g. H₃PO₄
• H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O.

Section 3 of 18Ionic equations for acid reactions

• Method
  • write all ions, then cancel the spectator ions to show H⁺(aq) as the active species

Acid with	Products	Ionic equation
Metal	Salt + hydrogen	2H+(aq) + Zn(s) → Zn2+(aq) + H2(g)
Metal oxide	Salt + water	2H+(aq) + CaO(s) → Ca2+(aq) + H2O(l)
Insoluble carbonate	Salt + water + CO2	2H+(aq) + CuCO3(s) → Cu2+(aq) + H2O(l) + CO2(g)
Soluble carbonate	Salt + water + CO2	2H+(aq) + CO32−(aq) → H2O(l) + CO2(g)
Alkali	Salt + water	H+(aq) + OH−(aq) → H2O(l)

Section 4 of 18Acid dissociation constant K_a and pK_a

Weak acid

Weak acid an acid that partially dissociates in aqueous solution

• Examples are organic acids such as ethanoic acid, and HCN, H₂S and H₂CO₃.
• The equilibrium lies mostly to the left.

Ka

[H⁺][A⁻] ÷ [HA], with units mol dm⁻³

What the symbols mean: K_a = acid dissociation constant (mol dm⁻³) · [H⁺], [A⁻], [HA] = equilibrium concentrations (mol dm⁻³); larger K_a = stronger acid

• Values are very small, e.g. 1.74 × 10⁻⁵ mol dm⁻³ for ethanoic acid.
• A higher K_a means more dissociation and a stronger acid.

pKa

−log K_a

• It makes the wide range of small K_a values easier to handle.
• Most weak acids have pK_a between 3 and 7.

Section 5 of 18pH and the ionic product of water

1pH

−log[H⁺],

• and [H⁺] = 10^−pH
  What the symbols mean: pH has no unit · [H⁺] = concentration of hydrogen ions (mol dm⁻³); each 1 pH unit is a ten-fold change in [H⁺]
• The scale is logarithmic, so pH 5 has ten times the [H⁺] of pH 6.
• pH values are usually given to 2 decimal places.

2Ionic product of water, Kw

[H⁺][OH⁻] = 1 × 10⁻¹⁴ mol² dm⁻⁶

What the symbols mean: K_w = ionic product of water (1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K) · [H⁺], [OH⁻] = ion concentrations (mol dm⁻³)

• It comes from the equilibrium in which a few water molecules dissociate into H⁺ and OH⁻.
• [OH⁻] = K_w ÷ [H⁺] and [H⁺] = K_w ÷ [OH⁻].

Section 6 of 18pH of strong acids and strong bases

Strong acid

• Strong acid fully ionised, HA → H⁺ + A⁻, so [H⁺] = [HA]
• H⁺ from water is negligible.
• 0.01 mol dm⁻³ HCl has pH = −log 0.01 = 2.00.

Strong base

• Strong base fully ionised, so [OH⁻] equals the concentration of base
• Find [H⁺] = K_w ÷ [OH⁻], then pH = −log[H⁺].
• 0.15 mol dm⁻³ NaOH gives [H⁺] = 6.66 × 10⁻¹⁴ and pH 13.17.

Finding [OH−] from pH

• Finding [OH−] from pH pH 10.50 gives [H⁺] = 10^−10.50 = 3.16 × 10⁻¹¹ mol dm⁻³.
• [OH⁻] = 10⁻¹⁴ ÷ 3.16 × 10⁻¹¹ = 3.16 × 10⁻⁴ mol dm⁻³.

Check

• [OH⁻] of 1.0 × 10⁻³ mol dm⁻³ gives [H⁺] = 1.0 × 10⁻¹¹, so pH 11.00

Section 7 of 18pH of a weak acid

Data needed

• Data needed the concentration of the acid and its K_a

Simplification

• HA dissociates into one H⁺ and one A⁻, so [H⁺] = [A⁻]
• K_a = [H⁺]² ÷ [HA], so [H⁺] = √(K_a × [HA]).

1

Write the K_a expression.

2

Set [H⁺] = [A⁻] and rearrange for [H⁺].

3

Substitute K_a and [HA].

4

Calculate pH = −log[H⁺].

• Example
  • 0.100 mol dm⁻³ ethanoic acid with K_a 1.74 × 10⁻⁵
  • [H⁺] = 1.32 × 10⁻³ mol dm⁻³, so pH 2.88.

Section 8 of 18Assumptions in weak acid calculations

Assumptions

• [H⁺] equals [A⁻] at equilibrium because of 1 : 1 dissociation and negligible H⁺ from water.
• Dissociation is so small that the equilibrium [HA] equals the initial [HA].

Acid strength

• Acid strength the stronger the acid, the higher [H⁺] at equilibrium and the larger K_a
• Dissociation is greater, so the difference between initial and equilibrium [HA] is greater and the approximation is less valid.

Section 9 of 18Buffer solutions

Buffer

Buffer a solution that resists changes in pH when small amounts of acid or alkali are added

• It keeps [H⁺] and [OH⁻] almost constant.

Acidic buffer

• Acidic buffer a weak acid and a soluble salt of the acid, e.g. ethanoic acid and sodium ethanoate
• CH₃COOH ⇌ H⁺ + CH₃COO⁻ is partial, so the acid has a high concentration.
• CH₃COONa fully dissociates, giving a high concentration of CH₃COO⁻.
• The mixture has large reservoirs of both the weak acid and its conjugate base.

Section 10 of 18How a buffer acts

Added H⁺

• The extra H⁺ reacts with CH₃COO⁻ to form CH₃COOH.
• The equilibrium shifts left, limiting the fall in pH.
• The large supply of CH₃COO⁻ means its concentration barely changes, so pH stays almost constant.

Added OH⁻

• OH⁻ reacts with H⁺ to form water.
• The fall in [H⁺] shifts the equilibrium right, so more CH₃COOH dissociates to replace H⁺.
• The pH therefore stays nearly constant.

Section 11 of 18Making a buffer from excess weak acid and strong alkali

Method

• Method mix a weak acid with a limited amount of strong alkali, with the weak acid in excess
• CH₃COOH + NaOH → CH₃COONa + H₂O uses up all the alkali.
• Excess CH₃COOH remains, together with the CH₃COONa formed.
• This is the same acid and conjugate base mixture, so it buffers in the same way.

Key condition

• Key condition the weak acid must be in excess; if the alkali were in excess no buffer would form

Explanation needed

• Explanation needed state that the weak acid is in excess and that both the weak acid and its conjugate base are present
  • (partial neutralisation)

Section 12 of 18Calculating the pH of a buffer

• Data needed
  • K_a and the concentrations of the weak acid and its salt
  • K_a = [salt][H⁺] ÷ [acid], so [H⁺] = K_a × [acid] ÷ [salt].
  • In logarithms, pH = pK_a + log([salt] ÷ [acid]), the Henderson–Hasselbalch equation.

1

Write the K_a expression.

2

Rearrange to [H⁺] = K_a × [acid] ÷ [salt].

3

Substitute the concentrations.

4

Calculate pH = −log[H⁺].

• Example
  • 0.305 mol dm⁻³ ethanoic acid and 0.520 mol dm⁻³ sodium ethanoate, K_a 1.74 × 10⁻⁵
  • [H⁺] = 1.74 × 10⁻⁵ × 0.305 ÷ 0.520 = 1.02 × 10⁻⁵ mol dm⁻³.
  • The pH is 4.99.

Section 13 of 18Blood buffer

Role of HCO₃⁻

• Role of HCO₃⁻ it keeps blood pH between 7.35 and 7.45
• Cells produce CO₂ in aerobic respiration, which forms H⁺ in blood: CO₂ + H₂O ⇌ H⁺ + HCO₃⁻.

Added H⁺

• Added H⁺ the equilibrium shifts left, lowering [H⁺] and keeping pH constant

Removed H⁺

• Removed H⁺ the equilibrium shifts right, raising [H⁺] and keeping pH constant

Importance

• Importance uncontrolled H⁺ would lower blood pH and cause acidosis
• Acidosis is too much acid in body fluids, which can cause malfunction and eventually coma.

Section 14 of 18pH titration curves

pH curve

pH curve a graph of pH against volume of acid or base added, measured with a pH meter

• It is S-shaped, and the midpoint of the vertical section is the equivalence (stoichiometric) point.

Information from a curve

• The starting pH, which gives the pH of the solution in the flask.
• The pH at the equivalence point.
• The volume of titrant at the equivalence point.
• The pH range of the vertical section.

Four combinations

• Four combinations strong acid + strong base, weak acid + strong base, strong acid + weak base, weak acid + weak base
• Identify which is shown from the starting and finishing pH.
• Plotting pH against volume of acid added gives the mirror image with the same information.

Section 15 of 18Acid–base indicators

Indicator

Indicator a weak acid HIn whose conjugate base In⁻ has a different colour

• HIn + H₂O ⇌ H₃O⁺ + In⁻.

Colour change

• Colour change it depends on the ratio [HIn] : [In⁻]
• In acid the equilibrium shifts left and the HIn colour dominates.
• In alkali it shifts right and the In⁻ colour dominates.
• The colour changes gradually over a pH range, not at a single pH.

End point

• [HIn] = [In⁻], so pK_a of the indicator = pH at its end point

Section 16 of 18Choosing an indicator

Choosing an indicator

Principle

• Principle the pH changes very rapidly around the equivalence point
• An indicator changes colour over a narrow range centred near its pK_a.
• It is suitable if its range falls within the rapid pH change of the titration.

Strong acid with strong base

• Strong acid with strong base the pH changes from about 4 to 10 at the end point
• Methyl red and phenolphthalein are both suitable.

Weak acid with strong base

• Weak acid with strong base the equivalence point is above pH 7, so phenolphthalein is suitable and methyl red is not

Strong acid with weak base

• Strong acid with weak base the equivalence point is below pH 7, so methyl red is suitable and phenolphthalein is not

Weak acid with weak base

• Weak acid with weak base there is no sharp vertical section, so no simple indicator is suitable

Section 17 of 18Key terms

Term	Meaning
Bronsted acid	A proton donor
Conjugate acid–base pair	Two species that differ by one H+
Ka	Acid dissociation constant, [H+][A−] ÷ [HA]
pH	−log of the hydrogen ion concentration in mol dm−3
Kw	Ionic product of water, 1 × 10−14 mol2 dm−6
Monobasic acid	An acid with one ionisable hydrogen per molecule

Term	Meaning
Buffer solution	A solution that resists changes in pH on adding small amounts of acid or alkali
Equivalence point	The point at which the acid and base have reacted in exact stoichiometric amounts
Indicator	A weak acid whose acid and conjugate base forms have different colours
Henderson–Hasselbalch equation	pH = pKa + log([salt] ÷ [acid])
Acidosis	A condition of too much acid in body fluids such as blood

Section 18 of 18Exam technique

1

Using [H⁺] = [HA] for a weak acid — only strong acids fully dissociate

2

Forgetting to take the square root when finding [H⁺] of a weak acid

3

Calculating pH of a strong base as −log[OH⁻] — find [H⁺] from K_w first

4

Confusing K_a and pK_a — a stronger acid has a larger K_a and a smaller pK_a

1

Saying a buffer stops pH changing — it only resists changes

2

Explaining a buffer without stating the weak acid is in excess and both acid and conjugate base are present

3

Using the wrong ratio in the buffer expression — [H⁺] = K_a × [acid] ÷ [salt]

4

Choosing an indicator whose range lies outside the vertical section of the curve