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Wyatt's Notes — IB

Acids and Bases

Bronsted-Lowry Theory

Definitions

  • Acid: A proton (H+\mathrm{H}^+) donor.
  • Base: A proton (H+\mathrm{H}^+) acceptor.

Conjugate Acid-Base Pairs

When an acid donates a proton, the remaining species is its conjugate base. When a base accepts A proton, the resulting species is its conjugate acid.

HA+H2O⇌H3O++A−\mathrm{HA} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}_3\mathrm{O}^+ + \mathrm{A}^-
Left sideRight side
HA (acid)A−^- (conjugate base)
H2_2O (base)H3_3O+^+ (conjugate acid)

Key relationship: The stronger the acid, the weaker its conjugate base, and vice versa.

Strong vs Weak Acids and Bases

PropertyStrong Acid/BaseWeak Acid/Base
DissociationComplete (100%)Partial (equilibrium)
EquilibriumLies far to the rightLies to the left
ConductivityHighLow
pH (same conc.)Lower (acid) / Higher (base)Less extreme
Reaction rate with metalsFasterSlower
Examples (acids)HCl, HNO3_3H2_2SO4_4HClO4_4CH3_3COOH, HF, HCN, H2_2CO3_3
Examples (bases)NaOH, KOH, Ba(OH)2_2NH3_3CH3_3COO−^-CO32−_3^{2-}

:::caution[Exam Tip] H2_2SO4_4 is a diprotic acid. The first dissociation is complete (strong), But the second dissociation is partial (weak): HSO4−⇌H++SO42−\mathrm{HSO}_4^- \rightleftharpoons \mathrm{H}^+ + \mathrm{SO}_4^{2-} with Ka≈1.0×10−2K_a \approx 1.0 \times 10^{-2}.

pH and pOH Scales

Definitions

pH=−log⁡[H+]\mathrm{pH} = -\log[\mathrm{H}^+] pOH=−log⁡[OH−]\mathrm{pOH} = -\log[\mathrm{OH}^-]

Relationship

At 25°C25\degree\mathrm{C}:

pH+pOH=14\mathrm{pH} + \mathrm{pOH} = 14 [H+][OH−]=Kw=1.0×10−14[\mathrm{H}^+][\mathrm{OH}^-] = K_w = 1.0 \times 10^{-14}

pH Scale

pHNature[H+^+] (mol/L)
0Strongly acidic1.0×1001.0 \times 10^0
11.0×10−11.0 \times 10^{-1}
31.0×10−31.0 \times 10^{-3}
7Neutral1.0×10−71.0 \times 10^{-7}
111.0×10−111.0 \times 10^{-11}
14Strongly basic1.0×10−141.0 \times 10^{-14}
Worked Example 1: pH of a Strong Acid

Calculate the pH of 0.050mol/L0.050\mathrm{ mol/L} HCl.

HCl is a strong acid and dissociates completely:

[H+]=0.050mol/L[\mathrm{H}^+] = 0.050\mathrm{ mol/L} pH=−log⁡(0.050)=−log⁡(5.0×10−2)=2−log⁡5.0=2−0.699=1.30\mathrm{pH} = -\log(0.050) = -\log(5.0 \times 10^{-2}) = 2 - \log 5.0 = 2 - 0.699 = 1.30
Worked Example 2: pH of a Strong Base

Calculate the pH of 0.020mol/L0.020\mathrm{ mol/L} NaOH at 25°C25\degree\mathrm{C}.

NaOH is a strong base:

[OH−]=0.020mol/L[\mathrm{OH}^-] = 0.020\mathrm{ mol/L} pOH=−log⁡(0.020)=−log⁡(2.0×10−2)=2−log⁡2.0=2−0.301=1.70\mathrm{pOH} = -\log(0.020) = -\log(2.0 \times 10^{-2}) = 2 - \log 2.0 = 2 - 0.301 = 1.70 pH=14−1.70=12.30\mathrm{pH} = 14 - 1.70 = 12.30
Worked Example 3: pH of a Diprotic Strong Acid

Calculate the pH of 0.010mol/L0.010\mathrm{ mol/L} H2_2SO4_4.

Since the first dissociation of H2_2SO4_4 is complete:

H2SO4→H++HSO4−\mathrm{H}_2\mathrm{SO}_4 \to \mathrm{H}^+ + \mathrm{HSO}_4^-

Each mole of H2_2SO4_4 gives 1 mole of H+^+ from the first dissociation. At this concentration, The second dissociation contributes additional H+^+But for most IB exam questions, it is Acceptable to consider only the first dissociation unless told otherwise:

[H+]≈0.010mol/L[\mathrm{H}^+] \approx 0.010\mathrm{ mol/L} pH=−log⁡(0.010)=2.00\mathrm{pH} = -\log(0.010) = 2.00

If the second dissociation is considered (with Ka2=1.0×10−2K_{a2} = 1.0 \times 10^{-2}):

Let xx be the additional [H+\mathrm{H}^+] from the second dissociation:

Ka2=[H+][SO42−][HSO4−]=(0.010+x)(x)0.010−x=1.0×10−2K_{a2} = \frac{[\mathrm{H}^+][\mathrm{SO}_4^{2-}]}{[\mathrm{HSO}_4^-]} = \frac{(0.010 + x)(x)}{0.010 - x} = 1.0 \times 10^{-2}

This gives x≈0.0045x \approx 0.0045So [H+]≈0.0145[\mathrm{H}^+] \approx 0.0145 and pH≈1.84\mathrm{pH} \approx 1.84.

Ka and Kb Expressions

Acid Dissociation Constant

For a weak acid HA\mathrm{HA}:

HA⇌H++A−\mathrm{HA} \rightleftharpoons \mathrm{H}^+ + \mathrm{A}^- Ka=[H+][A−][HA]K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]}

Base Dissociation Constant

For a weak base B\mathrm{B}:

B+H2O⇌BH++OH−\mathrm{B} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{BH}^+ + \mathrm{OH}^- Kb=[BH+][OH−][B]K_b = \frac{[\mathrm{BH}^+][\mathrm{OH}^-]}{[\mathrm{B}]}

Relationship Between Ka and Kb

For a conjugate acid-base pair:

Ka×Kb=Kw=1.0×10−14K_a \times K_b = K_w = 1.0 \times 10^{-14} pKa+pKb=14\mathrm{p}K_a + \mathrm{p}K_b = 14

Where pKa=−log⁡Ka\mathrm{p}K_a = -\log K_a.

Common Ka Values at 25°C25\degree\mathrm{C}

AcidKapKa
CH3_3COOH1.8×10−51.8 \times 10^{-5}4.74
HCOOH (formic)1.8×10−41.8 \times 10^{-4}3.74
HCN6.2×10−106.2 \times 10^{-10}9.21
HF6.8×10−46.8 \times 10^{-4}3.17
HNO2_24.5×10−44.5 \times 10^{-4}3.35
NH4+_4^+5.6×10−105.6 \times 10^{-10}9.25
Worked Example 4: pH of a Weak Acid

Calculate the pH of 0.100mol/L0.100\mathrm{ mol/L} CH3_3COOH (Ka=1.8×10−5K_a = 1.8 \times 10^{-5}).

CH3COOH⇌H++CH3COO−\mathrm{CH}_3\mathrm{COOH} \rightleftharpoons \mathrm{H}^+ + \mathrm{CH}_3\mathrm{COO}^-

ICE table (initial, change, equilibrium):

Species[Initial]Change[Equilibrium]
CH3_3COOH0.100−x-x0.100−x0.100 - x
H+^+0+x+xxx
CH3_3COO−^-0+x+xxx
Ka=x20.100−x=1.8×10−5K_a = \frac{x^2}{0.100 - x} = 1.8 \times 10^{-5}

Check the 5%5\% rule: Kac=1.8×10−50.100=1.8×10−4<0.05\frac{K_a}{c} = \frac{1.8 \times 10^{-5}}{0.100} = 1.8 \times 10^{-4} \lt 0.05. The approximation 0.100−x≈0.1000.100 - x \approx 0.100 is valid.

X2=1.8×10−6X^2 = 1.8 \times 10^{-6} X=1.8×10−6=1.34×10−3mol/LX = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3}\mathrm{ mol/L} pH=−log⁡(1.34×10−3)=2.87\mathrm{pH} = -\log(1.34 \times 10^{-3}) = 2.87
Worked Example 5: pH of a Weak Base

Calculate the pH of 0.150mol/L0.150\mathrm{ mol/L} NH3_3 (Kb=1.8×10−5K_b = 1.8 \times 10^{-5}) at 25°C25\degree\mathrm{C}.

NH3+H2O⇌NH4++OH−\mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{NH}_4^+ + \mathrm{OH}^-
Species[Initial]Change[Equilibrium]
NH3_30.150−x-x0.150−x0.150 - x
NH4+_4^+0+x+xxx
OH−^-0+x+xxx
Kb=x20.150−x=1.8×10−5K_b = \frac{x^2}{0.150 - x} = 1.8 \times 10^{-5}

Approximation valid: Kbc=1.2×10−4<0.05\frac{K_b}{c} = 1.2 \times 10^{-4} \lt 0.05.

X2=0.150×1.8×10−5=2.7×10−6X^2 = 0.150 \times 1.8 \times 10^{-5} = 2.7 \times 10^{-6} X=2.7×10−6=1.64×10−3mol/L=[OH−]X = \sqrt{2.7 \times 10^{-6}} = 1.64 \times 10^{-3}\mathrm{ mol/L} = [\mathrm{OH}^-] pOH=−log⁡(1.64×10−3)=2.78\mathrm{pOH} = -\log(1.64 \times 10^{-3}) = 2.78 pH=14−2.78=11.22\mathrm{pH} = 14 - 2.78 = 11.22

Kw (Ionic Product of Water)

Water undergoes autoionisation:

H2O(l)⇌H+(aq)+OH−(aq)\mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \mathrm{H}^+(aq) + \mathrm{OH}^-(aq) Kw=[H+][OH−]K_w = [\mathrm{H}^+][\mathrm{OH}^-]

At 25°C25\degree\mathrm{C}: Kw=1.0×10−14K_w = 1.0 \times 10^{-14} (mol2^2/L2^2).

KwK_w is temperature dependent:

Temperature (°\degreeC)KwK_w
01.14×10−151.14 \times 10^{-15}
251.00×10−141.00 \times 10^{-14}
505.48×10−145.48 \times 10^{-14}
1005.13×10−135.13 \times 10^{-13}

:::caution[Exam Tip] At 50°C50\degree\mathrm{C}Pure water has pH=6.63\mathrm{pH} = 6.63 (not 7). This is Because KwK_w is larger, so [H+]=[OH−]=Kw>10−7[\mathrm{H}^+] = [\mathrm{OH}^-] = \sqrt{K_w} \gt 10^{-7}. The water Is still neutral because [H+]=[OH−][\mathrm{H}^+] = [\mathrm{OH}^-]. Neutral does not always mean pH = 7; on temperature.

Polyprotic Acids

A polyprotic acid can donate more than one proton. Each dissociation has its own KaK_a value.

Carbonic acid (H2_2CO3_3):

H2CO3⇌H++HCO3−Ka1=4.3×10−7\mathrm{H}_2\mathrm{CO}_3 \rightleftharpoons \mathrm{H}^+ + \mathrm{HCO}_3^- \quad K_{a1} = 4.3 \times 10^{-7} HCO3−⇌H++CO32−Ka2=4.8×10−11\mathrm{HCO}_3^- \rightleftharpoons \mathrm{H}^+ + \mathrm{CO}_3^{2-} \quad K_{a2} = 4.8 \times 10^{-11}

Phosphoric acid (H3_3PO4_4):

H3PO4⇌H++H2PO4−Ka1=7.5×10−3\mathrm{H}_3\mathrm{PO}_4 \rightleftharpoons \mathrm{H}^+ + \mathrm{H}_2\mathrm{PO}_4^- \quad K_{a1} = 7.5 \times 10^{-3} H2PO4−⇌H++HPO42−Ka2=6.2×10−8\mathrm{H}_2\mathrm{PO}_4^- \rightleftharpoons \mathrm{H}^+ + \mathrm{HPO}_4^{2-} \quad K_{a2} = 6.2 \times 10^{-8} HPO42−⇌H++PO43−Ka3=4.2×10−13\mathrm{HPO}_4^{2-} \rightleftharpoons \mathrm{H}^+ + \mathrm{PO}_4^{3-} \quad K_{a3} = 4.2 \times 10^{-13}

General rule: Ka1≫Ka2≫Ka3K_{a1} \gg K_{a2} \gg K_{a3}. Each successive proton is harder to remove because Removing a proton from an increasingly negative ion requires more energy.

Indicators

Common Acid-Base Indicators

IndicatorColour (Acid)Transition pH RangeColour (Base)
Methyl orangeRed3.1 — 4.4Yellow
Bromothymol blueYellow6.0 — 7.6Blue
PhenolphthaleinColourless8.3 — 10.0Pink/Magenta
Universal indicatorRed (pH 1)Full range 1—14Violet (pH 14)

How Indicators Work

Indicators are weak organic acids (HIn) where the acid form and conjugate base form have different Colours:

HIn⇌H++In−\mathrm{HIn} \rightleftharpoons \mathrm{H}^+ + \mathrm{In}^-

The colour observed depends on the ratio [HIn]/[In−][\mathrm{HIn}]/[\mathrm{In}^-]Which depends on [H+^+] (pH).

Choosing an Indicator for a Titration

An indicator is suitable when its transition range overlaps with the equivalence point of the Titration (where pH changes most steeply).

Titration TypeEquivalence Point pHSuitable Indicator
Strong acid — strong basepH = 7Bromothymol blue, phenolphthalein, methyl orange
Weak acid — strong basepH \gt 7Phenolphthalein
Strong acid — weak basepH \lt 7Methyl orange

pH Curves for Titrations

Strong Acid — Strong Base

Example: HCl + NaOH

  • Initial pH: low (strong acid).
  • pH rises slowly, then sharply near the equivalence point (pH = 7).
  • Beyond equivalence point: pH approaches 14 (excess strong base).

Weak Acid — Strong Base

Example: CH3_3COOH + NaOH

  • Initial pH: higher than strong acid at same concentration (partial dissociation).
  • Buffer region: gradual pH rise before the equivalence point.
  • Equivalence point: pH \gt 7 (conjugate base of weak acid hydrolyses).
  • Suitable indicator: phenolphthalein.

Strong Acid — Weak Base

Example: HCl + NH3_3

  • Initial pH: low (strong acid).
  • Equivalence point: pH \lt 7 (conjugate acid of weak base hydrolyses).
  • Suitable indicator: methyl orange.

Key Features of pH Curves

  • Equivalence point: The volume where stoichiometrically equivalent amounts of acid and base have reacted. The pH at this point depends on the salt formed.
  • Half-equivalence point: The volume where half the acid/base has been neutralised. At this point, pH=pKa\mathrm{pH} = \mathrm{p}K_a (for weak acid titrations), because [HA]=[A−][\mathrm{HA}] = [\mathrm{A}^-].

Buffer Solutions

Composition

A buffer solution resists changes in pH when small amounts of acid or base are added. It Consists of:

  1. A weak acid and its conjugate base, or
  2. A weak base and its conjugate acid.

Examples:

  • CH3_3COOH / CH3_3COONa−^- (acetic acid / acetate)
  • NH3_3 / NH4_4Cl (ammonia / ammonium)
  • H2_2CO3_3 / HCO3−_3^- (carbonic acid / bicarbonate — the blood buffer system)

Mechanism

Adding acid (H+^+): The conjugate base (A−^-) reacts with the added H+^+:

A−+H+→HA\mathrm{A}^- + \mathrm{H}^+ \to \mathrm{HA}

Adding base (OH−^-): The weak acid (HA) reacts with the added OH−^-:

HA+OH−→A−+H2O\mathrm{HA} + \mathrm{OH}^- \to \mathrm{A}^- + \mathrm{H}_2\mathrm{O}

In both cases, the ratio [HA]/[A−][\mathrm{HA}]/[\mathrm{A}^-] changes only slightly, so pH remains nearly Constant.

Henderson-Hasselbalch Equation

pH=pKa+log⁡[A−][HA]\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}

Or equivalently:

pH=pKa+log⁡[base][acid]\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\mathrm{base}]}{[\mathrm{acid}]}

Buffer Capacity

Buffer capacity is the amount of acid or base that can be added before the pH changes significantly. A buffer is most effective when:

  • The concentrations of the weak acid and conjugate base are large (more moles available to react).
  • The ratio [A−]/[HA][\mathrm{A}^-]/[\mathrm{HA}] is close to 1 (i.e., pH is close to pKaK_a).

The effective buffer range is approximately pKa±1\mathrm{p}K_a \pm 1.

Worked Example 6: Buffer pH Calculation

Calculate the pH of a buffer solution containing 0.200mol/L0.200\mathrm{ mol/L} CH3_3COOH and 0.150mol/L0.150\mathrm{ mol/L} CH3_3COONa. (Ka=1.8×10−5K_a = 1.8 \times 10^{-5}PKa=4.74K_a = 4.74)

Using the Henderson-Hasselbalch equation:

pH=4.74+log⁡0.1500.200\mathrm{pH} = 4.74 + \log\frac{0.150}{0.200} pH=4.74+log⁡(0.750)\mathrm{pH} = 4.74 + \log(0.750) pH=4.74+(−0.125)=4.62\mathrm{pH} = 4.74 + (-0.125) = 4.62
Worked Example 7: pH Change When Adding Acid to a Buffer

To 500mL500\mathrm{ mL} of the buffer from Worked Example 6, 0.010mol0.010\mathrm{ mol} of HCl is added. Calculate the new pH.

Initial moles:

  • n(CH3COOH)=0.200×0.500=0.100moln(\mathrm{CH}_3\mathrm{COOH}) = 0.200 \times 0.500 = 0.100\mathrm{ mol}
  • n(CH3COO−)=0.150×0.500=0.075moln(\mathrm{CH}_3\mathrm{COO}^-) = 0.150 \times 0.500 = 0.075\mathrm{ mol}

Adding 0.010mol0.010\mathrm{ mol} HCl:

  • HCl reacts with CH3_3COO−^-: CH3COO−+H+→CH3COOH\mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}^+ \to \mathrm{CH}_3\mathrm{COOH}
  • n(CH3COO−)=0.075−0.010=0.065moln(\mathrm{CH}_3\mathrm{COO}^-) = 0.075 - 0.010 = 0.065\mathrm{ mol}
  • n(CH3COOH)=0.100+0.010=0.110moln(\mathrm{CH}_3\mathrm{COOH}) = 0.100 + 0.010 = 0.110\mathrm{ mol}
pH=4.74+log⁡0.0650.110\mathrm{pH} = 4.74 + \log\frac{0.065}{0.110} pH=4.74+log⁡(0.591)=4.74+(−0.228)=4.51\mathrm{pH} = 4.74 + \log(0.591) = 4.74 + (-0.228) = 4.51

The pH changed from 4.62 to 4.51, a change of only 0.11 units, demonstrating the buffer”s Effectiveness.

Worked Example 8: Preparing a Buffer

What mass of sodium acetate (CH3_3COONa, Mr=82.0g/molM_r = 82.0\mathrm{ g/mol}) must be added to 1.00L1.00\mathrm{ L} of 0.100mol/L0.100\mathrm{ mol/L} CH3_3COOH to produce a buffer with pH = 5.00? (Ka=1.8×10−5K_a = 1.8 \times 10^{-5})

Using the Henderson-Hasselbalch equation:

5.00=4.74+log⁡[CH3COO−]0.1005.00 = 4.74 + \log\frac{[\mathrm{CH}_3\mathrm{COO}^-]}{0.100} log⁡[CH3COO−]0.100=0.26\log\frac{[\mathrm{CH}_3\mathrm{COO}^-]}{0.100} = 0.26 [CH3COO−]0.100=100.26=1.82\frac{[\mathrm{CH}_3\mathrm{COO}^-]}{0.100} = 10^{0.26} = 1.82 [CH3COO−]=0.182mol/L[\mathrm{CH}_3\mathrm{COO}^-] = 0.182\mathrm{ mol/L}

In 1.00L1.00\mathrm{ L}:

N(CH3COONa)=0.182molN(\mathrm{CH}_3\mathrm{COONa}) = 0.182\mathrm{ mol} M=n×Mr=0.182×82.0=14.9gM = n \times M_r = 0.182 \times 82.0 = 14.9\mathrm{ g}

Acid-Base Titrations

Titration Calculations

At the equivalence point, moles of acid = moles of base.

Strong acid — strong base:

H++OH−→H2O\mathrm{H}^+ + \mathrm{OH}^- \to \mathrm{H}_2\mathrm{O}

PH at equivalence point = 7 (neutral salt).

Weak acid — strong base:

HA+OH−→A−+H2O\mathrm{HA} + \mathrm{OH}^- \to \mathrm{A}^- + \mathrm{H}_2\mathrm{O}

The conjugate base A−^- hydrolyses:

A−+H2O⇌HA+OH−\mathrm{A}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{HA} + \mathrm{OH}^-

PH at equivalence point \gt 7.

Strong acid — weak base:

H++B→BH+\mathrm{H}^+ + \mathrm{B} \to \mathrm{BH}^+

The conjugate acid BH+^+ hydrolyses:

BH++H2O⇌B+H3O+\mathrm{BH}^+ + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{B} + \mathrm{H}_3\mathrm{O}^+

PH at equivalence point \lt 7.

Worked Example 9: Titration of Weak Acid with Strong Base

25.0mL25.0\mathrm{ mL} of 0.100mol/L0.100\mathrm{ mol/L} CH3_3COOH (Ka=1.8×10−5K_a = 1.8 \times 10^{-5}) is titrated With 0.100mol/L0.100\mathrm{ mol/L} NaOH. Calculate the pH at the equivalence point.

At the equivalence point, moles of NaOH = moles of CH3_3COOH:

N=0.100×0.0250=0.00250molN = 0.100 \times 0.0250 = 0.00250\mathrm{ mol}

Volume of NaOH required:

V=0.002500.100=0.0250L=25.0mLV = \frac{0.00250}{0.100} = 0.0250\mathrm{ L} = 25.0\mathrm{ mL}

Total volume at equivalence point: 25.0+25.0=50.0mL=0.0500L25.0 + 25.0 = 50.0\mathrm{ mL} = 0.0500\mathrm{ L}.

All CH3_3COOH has been converted to CH3_3COO−^-:

[CH3COO−]=0.002500.0500=0.0500mol/L[\mathrm{CH}_3\mathrm{COO}^-] = \frac{0.00250}{0.0500} = 0.0500\mathrm{ mol/L}

The acetate ion hydrolyses:

Kb=KwKa=1.0×10−141.8×10−5=5.56×10−10K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10} [OH−]=Kb×[CH3COO−]=5.56×10−10×0.0500[\mathrm{OH}^-] = \sqrt{K_b \times [\mathrm{CH}_3\mathrm{COO}^-]} = \sqrt{5.56 \times 10^{-10} \times 0.0500} [OH−]=2.78×10−11=5.27×10−6mol/L[\mathrm{OH}^-] = \sqrt{2.78 \times 10^{-11}} = 5.27 \times 10^{-6}\mathrm{ mol/L} pOH=−log⁡(5.27×10−6)=5.28\mathrm{pOH} = -\log(5.27 \times 10^{-6}) = 5.28 pH=14−5.28=8.72\mathrm{pH} = 14 - 5.28 = 8.72
Worked Example 10: pH at Half-Equivalence Point

Using the titration from Worked Example 9, calculate the pH when 12.5mL12.5\mathrm{ mL} of NaOH has been Added (half-equivalence point).

At half-equivalence point, half the acid has been neutralised:

N(NaOH)=0.100×0.0125=0.00125molN(\mathrm{NaOH}) = 0.100 \times 0.0125 = 0.00125\mathrm{ mol}

Moles of CH3_3COOH remaining: 0.00250−0.00125=0.00125mol0.00250 - 0.00125 = 0.00125\mathrm{ mol}

Moles of CH3_3COO−^- formed: 0.00125mol0.00125\mathrm{ mol}

Since [HA]=[A−][\mathrm{HA}] = [\mathrm{A}^-]:

pH=pKa=−log⁡(1.8×10−5)=4.74\mathrm{pH} = \mathrm{p}K_a = -\log(1.8 \times 10^{-5}) = 4.74

This is a general result: at the half-equivalence point, pH=pKa\mathrm{pH} = \mathrm{p}K_a.

Common Pitfalls

  1. Strong vs weak acid pH: A 0.10mol/L0.10\mathrm{ mol/L} strong acid has pH = 1.0, but a 0.10mol/L0.10\mathrm{ mol/L} weak acid has pH \gt 1.0 ( 2—3) because it only partially dissociates. Do not assume [H+]=c[\mathrm{H}^+] = c for weak acids.

  2. Diprotic acid contribution: H2_2SO4_4 gives 2 H+^+ per molecule, but H2_2CO3_3 does not give 2 H+^+ at normal concentrations because Ka2K_{a2} is very small. Only the first dissociation contributes significantly.

  3. pH + pOH = 14 only at 25°C25\degree\mathrm{C}: At other temperatures, use the actual KwK_w value. pH+pOH=pKw\mathrm{pH} + \mathrm{pOH} = \mathrm{p}K_w.

  4. Buffer range: A buffer is only effective within ±1\pm 1 pH unit of its pKaK_a. Outside this range, the buffer capacity is essentially zero.

  5. Equivalence point pH: For weak acid — strong base titrations, the equivalence point pH is \gt 7, not 7. For strong acid — weak base, it is \lt 7.

  6. Henderson-Hasselbalch validity: The equation assumes that the concentrations of HA and A−^- are much larger than [H+][\mathrm{H}^+] and [OH−][\mathrm{OH}^-]. It is not valid for very dilute solutions.

  7. Neutral pH: Neutral means [H+]=[OH−][\mathrm{H}^+] = [\mathrm{OH}^-]Which equals pH = 7 only at 25°C25\degree\mathrm{C}. At 50°C50\degree\mathrm{C}Neutral pH is approximately 6.63.

  8. KaK_a and KbK_b relationship: Remember Ka×Kb=KwK_a \times K_b = K_w. This connects a conjugate acid-base pair. The conjugate base of a weak acid has a calculable KbK_b.

  9. Titration indicator choice: Methyl orange (3.1—4.4) is suitable for strong acid — weak base titrations (equivalence pH \lt 7). Phenolphthalein (8.3—10.0) is suitable for weak acid — strong base (equivalence pH \gt 7). Do not use methyl orange for weak acid — strong base.

  10. Units in Ka/Kb: Concentrations in KaK_a and KbK_b expressions are in mol/L. The units of KaK_a and KbK_b are mol/L. KwK_w has units of mol2^2/L2^2.

Practice Problems

Question 1: Strong Acid/Base pH

(a) Calculate the pH of 0.0030mol/L0.0030\mathrm{ mol/L} HNO3_3.

(b) Calculate the pH of 0.0025mol/L0.0025\mathrm{ mol/L} Ca(OH)2_2 at 25°C25\degree\mathrm{C}.

(c) A solution has pH = 3.40. What is [H+][\mathrm{H}^+]?

Answer:

(a) HNO3_3 is a strong monoprotic acid:

[H+]=0.0030mol/L[\mathrm{H}^+] = 0.0030\mathrm{ mol/L} pH=−log⁡(3.0×10−3)=3−log⁡3.0=3−0.477=2.52\mathrm{pH} = -\log(3.0 \times 10^{-3}) = 3 - \log 3.0 = 3 - 0.477 = 2.52

(b) Ca(OH)2_2 is a strong base giving 2 OH−^- per formula unit:

[OH−]=2×0.0025=0.0050mol/L[\mathrm{OH}^-] = 2 \times 0.0025 = 0.0050\mathrm{ mol/L} pOH=−log⁡(5.0×10−3)=3−log⁡5.0=3−0.699=2.30\mathrm{pOH} = -\log(5.0 \times 10^{-3}) = 3 - \log 5.0 = 3 - 0.699 = 2.30 pH=14−2.30=11.70\mathrm{pH} = 14 - 2.30 = 11.70

(c)

[H+]=10−pH=10−3.40=3.98×10−4mol/L[\mathrm{H}^+] = 10^{-\mathrm{pH}} = 10^{-3.40} = 3.98 \times 10^{-4}\mathrm{ mol/L}
Question 2: Weak Acid pH and Degree of Dissociation

Hypochlorous acid (HOCl) has Ka=3.5×10−8K_a = 3.5 \times 10^{-8}.

(a) Calculate the pH of 0.050mol/L0.050\mathrm{ mol/L} HOCl.

(b) Calculate the percentage dissociation of HOCl at this concentration.

Answer:

(a)

HOCl⇌H++OCl−\mathrm{HOCl} \rightleftharpoons \mathrm{H}^+ + \mathrm{OCl}^- Ka=x20.050=3.5×10−8K_a = \frac{x^2}{0.050} = 3.5 \times 10^{-8} X2=1.75×10−9X^2 = 1.75 \times 10^{-9} X=4.18×10−5mol/LX = 4.18 \times 10^{-5}\mathrm{ mol/L} pH=−log⁡(4.18×10−5)=4.38\mathrm{pH} = -\log(4.18 \times 10^{-5}) = 4.38

(b)

Percentagedissociation=4.18×10−50.050×100%=0.084%\mathrm{Percentage dissociation} = \frac{4.18 \times 10^{-5}}{0.050} \times 100\% = 0.084\%
Question 3: Conjugate Pairs and Ka/Kb

(a) The KbK_b of NH3_3 is 1.8×10−51.8 \times 10^{-5}. Calculate KaK_a for NH4+_4^+.

(b) Is NH4+_4^+ acidic, basic, or neutral in aqueous solution? Explain.

Answer:

(a)

Ka(NH4+)=KwKb(NH3)=1.0×10−141.8×10−5=5.56×10−10K_a(\mathrm{NH}_4^+) = \frac{K_w}{K_b(\mathrm{NH}_3)} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}

(b) NH4+_4^+ is the conjugate acid of the weak base NH3_3. Since NH3_3 is a weak base, its Conjugate acid NH4+_4^+ will donate protons in water, making the solution acidic. This is confirmed By the relatively large KaK_a value (5.56×10−10≫Kb5.56 \times 10^{-10} \gg K_b of NH4+_4^+ which would be Kw/Ka=1.8×10−5K_w/K_a = 1.8 \times 10^{-5}But wait — we already have KaK_a for NH4+_4^+So we can see it is An acid). A 0.1mol/L0.1\mathrm{ mol/L} NH4_4Cl solution would have pH \lt 7.

Question 4: Buffer Preparation and pH Change

A buffer is prepared by mixing 0.150mol/L0.150\mathrm{ mol/L} HCOOH (Ka=1.8×10−4K_a = 1.8 \times 10^{-4} PKa=3.74K_a = 3.74) with 0.100mol/L0.100\mathrm{ mol/L} HCOONa in equal volumes.

(a) Calculate the pH of the buffer.

(b) To 100mL100\mathrm{ mL} of this buffer, 5.0mL5.0\mathrm{ mL} of 0.100mol/L0.100\mathrm{ mol/L} NaOH is added. Calculate the new pH.

Answer:

(a) Equal volumes of 0.1500.150 and 0.100mol/L0.100\mathrm{ mol/L} give concentrations of 0.07500.0750 and 0.0500mol/L0.0500\mathrm{ mol/L} respectively:

pH=3.74+log⁡0.05000.0750=3.74+log⁡(0.667)=3.74+(−0.176)=3.56\mathrm{pH} = 3.74 + \log\frac{0.0500}{0.0750} = 3.74 + \log(0.667) = 3.74 + (-0.176) = 3.56

(b) Moles in 100mL100\mathrm{ mL} of buffer:

  • n(HCOOH)=0.0750×0.100=0.00750moln(\mathrm{HCOOH}) = 0.0750 \times 0.100 = 0.00750\mathrm{ mol}
  • n(HCOO−)=0.0500×0.100=0.00500moln(\mathrm{HCOO}^-) = 0.0500 \times 0.100 = 0.00500\mathrm{ mol}

Moles of NaOH added: n=0.100×0.0050=0.000500moln = 0.100 \times 0.0050 = 0.000500\mathrm{ mol}

After reaction:

  • n(HCOOH)=0.00750−0.000500=0.00700moln(\mathrm{HCOOH}) = 0.00750 - 0.000500 = 0.00700\mathrm{ mol}
  • n(HCOO−)=0.00500+0.000500=0.00550moln(\mathrm{HCOO}^-) = 0.00500 + 0.000500 = 0.00550\mathrm{ mol}
pH=3.74+log⁡0.005500.00700=3.74+log⁡(0.786)=3.74+(−0.105)=3.64\mathrm{pH} = 3.74 + \log\frac{0.00550}{0.00700} = 3.74 + \log(0.786) = 3.74 + (-0.105) = 3.64
Question 5: Titration Curve Analysis (Paper 2 Style)

20.0mL20.0\mathrm{ mL} of 0.100mol/L0.100\mathrm{ mol/L} NH3_3 (Kb=1.8×10−5K_b = 1.8 \times 10^{-5}) is titrated with 0.100mol/L0.100\mathrm{ mol/L} HCl.

(a) Calculate the pH at the equivalence point.

(b) State and explain which indicator would be most suitable for this titration.

(c) Calculate the pH when 10.0mL10.0\mathrm{ mL} of HCl has been added (half-equivalence point).

Answer:

(a) At equivalence point, moles of HCl = moles of NH3_3:

N=0.100×0.0200=0.00200molN = 0.100 \times 0.0200 = 0.00200\mathrm{ mol}

Volume of HCl: V=0.00200/0.100=0.0200L=20.0mLV = 0.00200 / 0.100 = 0.0200\mathrm{ L} = 20.0\mathrm{ mL}.

Total volume: 20.0+20.0=40.0mL=0.0400L20.0 + 20.0 = 40.0\mathrm{ mL} = 0.0400\mathrm{ L}.

All NH3_3 is converted to NH4+_4^+:

[NH4+]=0.002000.0400=0.0500mol/L[\mathrm{NH}_4^+] = \frac{0.00200}{0.0400} = 0.0500\mathrm{ mol/L} Ka(NH4+)=KwKb=1.0×10−141.8×10−5=5.56×10−10K_a(\mathrm{NH}_4^+) = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10} [H+]=Ka×[NH4+]=5.56×10−10×0.0500[\mathrm{H}^+] = \sqrt{K_a \times [\mathrm{NH}_4^+]} = \sqrt{5.56 \times 10^{-10} \times 0.0500} [H+]=2.78×10−11=5.27×10−6mol/L[\mathrm{H}^+] = \sqrt{2.78 \times 10^{-11}} = 5.27 \times 10^{-6}\mathrm{ mol/L} pH=−log⁡(5.27×10−6)=5.28\mathrm{pH} = -\log(5.27 \times 10^{-6}) = 5.28

(b) The equivalence point pH is 5.28, which falls within the transition range of methyl orange (3.1—4.4) only partially. A better choice would be bromocresol green (3.8—5.4) or methyl Red (4.4—6.2), which has a transition range that includes pH 5.28. From the common indicators Listed in the IB syllabus, methyl orange is the closest suitable indicator for a strong acid — weak Base titration.

(c) At the half-equivalence point, [NH3]=[NH4+][\mathrm{NH}_3] = [\mathrm{NH}_4^+]So:

pH=pKa(NH4+)=14−pKb=14−(−log⁡(1.8×10−5))\mathrm{pH} = \mathrm{p}K_a(\mathrm{NH}_4^+) = 14 - \mathrm{p}K_b = 14 - (-\log(1.8 \times 10^{-5})) pH=14−4.74=9.26\mathrm{pH} = 14 - 4.74 = 9.26
Question 6: Polyprotic Acid (Paper 2 Style)

Calculate the pH of a 0.100mol/L0.100\mathrm{ mol/L} H3_3PO4_4 solution. Use the following data: K_{a1} = 7.5 \times 10^{-3}$$K_{a2} = 6.2 \times 10^{-8}$$K_{a3} = 4.2 \times 10^{-13}.

Answer:

For the first dissociation:

H3PO4⇌H++H2PO4−\mathrm{H}_3\mathrm{PO}_4 \rightleftharpoons \mathrm{H}^+ + \mathrm{H}_2\mathrm{PO}_4^-

Since Ka1K_{a1} is not very small compared to ccThe approximation c−x≈cc - x \approx c may not be Valid. Check: Ka1/c=7.5×10−3/0.100=0.075>0.05K_{a1}/c = 7.5 \times 10^{-3}/0.100 = 0.075 \gt 0.05. The 5%5\% rule fails.

Solve the quadratic: x2+Kax−Ka⋅c=0x^2 + K_a x - K_a \cdot c = 0

X2+7.5×10−3x−7.5×10−4=0X^2 + 7.5 \times 10^{-3}x - 7.5 \times 10^{-4} = 0 X=−7.5×10−3+(7.5×10−3)2+4×7.5×10−42X = \frac{-7.5 \times 10^{-3} + \sqrt{(7.5 \times 10^{-3})^2 + 4 \times 7.5 \times 10^{-4}}}{2} X=−7.5×10−3+5.625×10−5+3.0×10−32X = \frac{-7.5 \times 10^{-3} + \sqrt{5.625 \times 10^{-5} + 3.0 \times 10^{-3}}}{2} X=−7.5×10−3+3.056×10−32X = \frac{-7.5 \times 10^{-3} + \sqrt{3.056 \times 10^{-3}}}{2} X=−7.5×10−3+5.528×10−22=4.778×10−22=2.389×10−2X = \frac{-7.5 \times 10^{-3} + 5.528 \times 10^{-2}}{2} = \frac{4.778 \times 10^{-2}}{2} = 2.389 \times 10^{-2} [H+]≈0.0239mol/L[\mathrm{H}^+] \approx 0.0239\mathrm{ mol/L} pH=−log⁡(0.0239)=1.62\mathrm{pH} = -\log(0.0239) = 1.62

Note: The second and third dissociations contribute negligible H+^+ since Ka2≪Ka1K_{a2} \ll K_{a1}.

Related Content at Other Levels

Summary

This topic covers the essential chemistry of acids and bases, including key reactions, underlying theories, and practical applications.

Key concepts include:

  • Brønsted-Lowry theory
  • strong and weak acids/bases
  • pH calculations
  • titration curves and indicators
  • hydrolysis of salts

Mastery of these concepts requires both theoretical understanding and the ability to apply knowledge to unfamiliar contexts, particularly in calculation and practical questions.

Cross-References

TopicSiteLink
[Acids and Bases]A-LevelView
[Acids and Bases]IBView
[Acids and Bases]DSEView

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.