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Wyatt's Notes — IB

Acids and Bases (Advanced)

1. Advanced pH Calculations

Strong Acids and Bases

For strong monoprotic acids, [H+][\mathrm{H}^+] equals the acid concentration. For strong diprotic Acids:

[H+]=2×[acid][\mathrm{H}^+] = 2 \times [\mathrm{acid}]

Example

0.050M0.050\mathrm{ M} H2SO4\mathrm{H}_2\mathrm{SO}_4 (assuming complete first dissociation and significant Second dissociation):

[H+]≈2×0.050=0.100M[\mathrm{H}^+] \approx 2 \times 0.050 = 0.100\mathrm{ M}pH=−log⁡(0.100)=1.00\mathrm{pH} = -\log(0.100) = 1.00

Weak Acids

For a weak acid HA\mathrm{HA} with acid dissociation constant KaK_a:

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

Assuming [H+]=[A−]=x[\mathrm{H}^+] = [\mathrm{A}^-] = x and [HA]≈c0[\mathrm{HA}] \approx c_0 (the 5% rule):

X=Ka⋅c0X = \sqrt{K_a \cdot c_0}pH=−log⁡x\mathrm{pH} = -\log x

Very Dilute Weak Acids

When the acid is very dilute (c0c_0 is small) or KaK_a is large, the approximation [HA]≈c0[\mathrm{HA}] \approx c_0 breaks down. Solve the full quadratic:

Ka=x2c0−xK_a = \frac{x^2}{c_0 - x}X2+Kax−Kac0=0X^2 + K_a x - K_a c_0 = 0

Example

Calculate the pH\mathrm{pH} of 0.010M0.010\mathrm{ M} ethanoic acid (Ka=1.8×10−5K_a = 1.8 \times 10^{-5}).

Using the approximation:

X=1.8×10−5×0.010=1.8×10−7=4.24×10−4X = \sqrt{1.8 \times 10^{-5} \times 0.010} = \sqrt{1.8 \times 10^{-7}} = 4.24 \times 10^{-4}

Check: 4.24×10−40.010×100%=4.24%<5%\dfrac{4.24 \times 10^{-4}}{0.010} \times 100\% = 4.24\% \lt 5\%. The approximation is Valid.

pH=−log⁡(4.24×10−4)=3.37\mathrm{pH} = -\log(4.24 \times 10^{-4}) = 3.37

Polyprotic Acids

For diprotic acids (H2A\mathrm{H}_2\mathrm{A}), the first dissociation dominates:

Ka1≫Ka2K_{a1} \gg K_{a2}

The pH\mathrm{pH} is determined primarily by Ka1K_{a1}. The second dissociation contributes Additional H+\mathrm{H}^+ but is negligible.

For H2SO4\mathrm{H}_2\mathrm{SO}_4: the first dissociation is complete (Ka1K_{a1} very large), but Ka2=1.2×10−2K_{a2} = 1.2 \times 10^{-2}.

Common Pitfalls

  • The 5% rule: if x/c0>5%x/c_0 \gt 5\%Use the quadratic formula.
  • For very dilute strong acids (c0<10−6Mc_0 \lt 10^{-6}\mathrm{ M}), the contribution of water”s autoionization ([H+]=10−7M[\mathrm{H}^+] = 10^{-7}\mathrm{ M}) becomes significant.
  • pOH=14−pH\mathrm{pOH} = 14 - \mathrm{pH} only holds at 25°C25\degree\mathrm{C}.

2. Buffer Solutions

Definition

A buffer solution resists changes in pH\mathrm{pH} when small amounts of acid or base are added. It consists of a weak acid and its conjugate base (or a weak base and its conjugate acid).

The Henderson-Hasselbalch Equation

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

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

Buffer Capacity

Buffer capacity is maximised when [HA]=[A−][\mathrm{HA}] = [\mathrm{A}^-]I.e., when pH=pKa\mathrm{pH} = \mathrm{p}K_a. A buffer is effective within ±1\pm 1 unit of its pKa\mathrm{p}K_a.

Preparing a Buffer

Example

Prepare an ethanoic acid/sodium ethanoate buffer with pH=5.00\mathrm{pH} = 5.00. Given pKa=4.76\mathrm{p}K_a = 4.76.

5.00=4.76+log⁡[CH3COO−][CH3COOH]5.00 = 4.76 + \log\frac{[\mathrm{CH}_3\mathrm{COO}^-]}{[\mathrm{CH}_3\mathrm{COOH}]}log⁡[CH3COO−][CH3COOH]=0.24\log\frac{[\mathrm{CH}_3\mathrm{COO}^-]}{[\mathrm{CH}_3\mathrm{COOH}]} = 0.24[CH3COO−][CH3COOH]=100.24=1.74\frac{[\mathrm{CH}_3\mathrm{COO}^-]}{[\mathrm{CH}_3\mathrm{COOH}]} = 10^{0.24} = 1.74

Mix ethanoic acid and sodium ethanoate in a molar ratio of 1:1.741 : 1.74 (or approximately 1:1.71 : 1.7).

Calculating pH Changes in a Buffer

When acid is added:

[HA]new=[HA]old+[H+]added[\mathrm{HA}]_{\mathrm{new}} = [\mathrm{HA}]_{\mathrm{old}} + [\mathrm{H}^+]_{\mathrm{added}}[A−]new=[A−]old−[H+]added[\mathrm{A}^-]_{\mathrm{new}} = [\mathrm{A}^-]_{\mathrm{old}} - [\mathrm{H}^+]_{\mathrm{added}}

When base is added:

[HA]new=[HA]old−[OH−]added[\mathrm{HA}]_{\mathrm{new}} = [\mathrm{HA}]_{\mathrm{old}} - [\mathrm{OH}^-]_{\mathrm{added}}[A−]new=[A−]old+[OH−]added[\mathrm{A}^-]_{\mathrm{new}} = [\mathrm{A}^-]_{\mathrm{old}} + [\mathrm{OH}^-]_{\mathrm{added}}

Example

A buffer contains 0.10M0.10\mathrm{ M} CH3COOH\mathrm{CH}_3\mathrm{COOH} and 0.10M0.10\mathrm{ M} CH3COO−\mathrm{CH}_3\mathrm{COO}^- (pH=4.76\mathrm{pH} = 4.76). Add 0.01mol0.01\mathrm{ mol} of HCl\mathrm{HCl} to 1.0L1.0\mathrm{ L} of buffer.

[CH3COOH]=0.10+0.01=0.11M[\mathrm{CH}_3\mathrm{COOH}] = 0.10 + 0.01 = 0.11\mathrm{ M}[CH3COO−]=0.10−0.01=0.09M[\mathrm{CH}_3\mathrm{COO}^-] = 0.10 - 0.01 = 0.09\mathrm{ M}pH=4.76+log⁡0.090.11=4.76+log⁡(0.818)=4.76−0.09=4.67\mathrm{pH} = 4.76 + \log\frac{0.09}{0.11} = 4.76 + \log(0.818) = 4.76 - 0.09 = 4.67

The pH\mathrm{pH} changed by only 0.090.09 units. Without the buffer, 0.01M0.01\mathrm{ M} HCl\mathrm{HCl} Would give pH=2.00\mathrm{pH} = 2.00.

Common Pitfalls

  • A buffer cannot withstand addition of large amounts of strong acid or base — it has finite capacity.
  • The Henderson-Hasselbalch equation assumes concentrations equal activities (valid for dilute solutions).
  • pH=pKa\mathrm{pH} = \mathrm{p}K_a does not mean the solution is neutral; it means the acid and conjugate base are present in equal amounts.

3. Indicator Theory

How Indicators Work

Acid-base indicators are weak acids (HIn\mathrm{HIn}) where the acid and conjugate base forms have Different colours:

HIn⇌H++In−KIn=[H+][In−][HIn]\mathrm{HIn} \rightleftharpoons \mathrm{H}^+ + \mathrm{In}^- \qquad K_{\mathrm{In}} = \frac{[\mathrm{H}^+][\mathrm{In}^-]}{[\mathrm{HIn}]}

The observed colour depends on the ratio [In−]/[HIn][\mathrm{In}^-]/[\mathrm{HIn}]:

ConditionDominant speciesColour observed
pH≪pKIn\mathrm{pH} \ll \mathrm{p}K_{\mathrm{In}}HIn\mathrm{HIn}Acid colour
pH≈pKIn\mathrm{pH} \approx \mathrm{p}K_{\mathrm{In}}Both presentTransition
pH≫pKIn\mathrm{pH} \gg \mathrm{p}K_{\mathrm{In}}In−\mathrm{In}^-Base colour

Effective Range

An indicator changes colour over approximately ±1\pm 1 unit around its pKIn\mathrm{p}K_{\mathrm{In}}:

pH range=pKIn±1\mathrm{pH\ range} = \mathrm{p}K_{\mathrm{In}} \pm 1

Choosing the Right Indicator

The indicator’s transition range should overlap with the equivalence point of the titration.

Titration typeEquivalence point pHSuitable indicator
Strong acid—strong basepH=7\mathrm{pH} = 7Bromothymol blue
Strong acid—weak basepH<7\mathrm{pH} \lt 7Methyl orange
Weak acid—strong basepH>7\mathrm{pH} \gt 7Phenolphthalein
Weak acid—weak baseNo sharp changeNo suitable indicator

Common Indicators

IndicatorpH\mathrm{pH} rangeAcid colourBase colour
Methyl orange3.13.1—4.44.4RedYellow
Bromothymol blue6.06.0—7.67.6YellowBlue
Phenolphthalein8.38.3—10.010.0ColourlessPink

4. Acid Deposition

Formation of Acid Rain

  1. Sulfur dioxide from combustion of fossil fuels is oxidised:
SO2+H2O→H2SO3\mathrm{SO}_2 + \mathrm{H}_2\mathrm{O} \to \mathrm{H}_2\mathrm{SO}_32SO2+O2→2SO3(catalysedbyV2O5)2\mathrm{SO}_2 + \mathrm{O}_2 \to 2\mathrm{SO}_3 \quad \mathrm{(catalysed by } \mathrm{V}_2\mathrm{O}_5)SO3+H2O→H2SO4\mathrm{SO}_3 + \mathrm{H}_2\mathrm{O} \to \mathrm{H}_2\mathrm{SO}_4
  1. Nitrogen oxides from high-temperature combustion:
N2+O2→2NO\mathrm{N}_2 + \mathrm{O}_2 \to 2\mathrm{NO}2NO+O2→2NO22\mathrm{NO} + \mathrm{O}_2 \to 2\mathrm{NO}_23NO2+H2O→2HNO3+NO3\mathrm{NO}_2 + \mathrm{H}_2\mathrm{O} \to 2\mathrm{HNO}_3 + \mathrm{NO}

Environmental Effects

EffectMechanism
Soil acidificationLeaches Ca2+\mathrm{Ca}^{2+}, Mg2+\mathrm{Mg}^{2+}Releases Al3+\mathrm{Al}^{3+}
Lake acidificationReduces biodiversity, mobilises toxic Al3+\mathrm{Al}^{3+}
Plant damageDamages leaves, inhibits root growth
Building corrosionCaCO3\mathrm{CaCO}_3 (limestone/marble) reacts with acid
Respiratory issuesFine sulfate/nitrate particles irritate lungs

Mitigation

  • Flue gas desulfurisation (FGD): SO2\mathrm{SO}_2 removed by reaction with CaO\mathrm{CaO} or CaCO3\mathrm{CaCO}_3.
  • Catalytic converters: reduce NOx\mathrm{NO}_x emissions.
  • Switching to low-sulfur fuels or renewable energy.

5. Solubility Product (KspK_{sp})

Definition

For a sparingly soluble salt MxAy\mathrm{M}_x\mathrm{A}_y:

MxAy(s)⇌xMy+(aq)+yAx−(aq)\mathrm{M}_x\mathrm{A}_y(s) \rightleftharpoons x\mathrm{M}^{y+}(aq) + y\mathrm{A}^{x-}(aq)K_`\{sp}` = [\mathrm{M}^{y+}]^x[\mathrm{A}^{x-}]^y

KspK_{sp} is the equilibrium constant for the dissolution of a solid. It is temperature-dependent.

Common KspK_{sp} Values

CompoundKspK_{sp}
AgCl\mathrm{AgCl}1.8×10−101.8 \times 10^{-10}
AgBr\mathrm{AgBr}5.0×10−135.0 \times 10^{-13}
AgI\mathrm{AgI}8.3×10−178.3 \times 10^{-17}
BaSO4\mathrm{BaSO}_41.1×10−101.1 \times 10^{-10}
CaCO3\mathrm{CaCO}_33.4×10−93.4 \times 10^{-9}
PbI2\mathrm{PbI}_27.1×10−97.1 \times 10^{-9}
Mg(OH)2\mathrm{Mg(OH)}_21.8×10−111.8 \times 10^{-11}

Solubility Calculations

Example

Calculate the solubility of AgCl\mathrm{AgCl} in g/L\mathrm{g/L}.

AgCl(s)⇌Ag+(aq)+Cl−(aq)\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^+(aq) + \mathrm{Cl}^-(aq)K_`\{sp}` = [\mathrm{Ag}^+][\mathrm{Cl}^-] = s^2S=1.8×10−10=1.3×10−5mol/LS = \sqrt{1.8 \times 10^{-10}} = 1.3 \times 10^{-5}\mathrm{ mol/L}Solubility=1.3×10−5×143.32=1.9×10−3g/L\mathrm{Solubility} = 1.3 \times 10^{-5} \times 143.32 = 1.9 \times 10^{-3}\mathrm{ g/L}

Common Ion Effect

Adding a common ion decreases the solubility of a sparingly soluble salt.

Example

Calculate the solubility of AgCl\mathrm{AgCl} in 0.10M0.10\mathrm{ M} NaCl\mathrm{NaCl}.

K_`\{sp}` = [\mathrm{Ag}^+][\mathrm{Cl}^-] = s \times 0.10S=1.8×10−100.10=1.8×10−9mol/LS = \frac{1.8 \times 10^{-10}}{0.10} = 1.8 \times 10^{-9}\mathrm{ mol/L}

Compared to pure water (1.3×10−5mol/L1.3 \times 10^{-5}\mathrm{ mol/L}), the solubility decreased by a factor Of about 70007000.

Precipitation

A precipitate forms when the ion product exceeds KspK_{sp}:

Q=[My+]x[Ax−]yQ = [\mathrm{M}^{y+}]^x[\mathrm{A}^{x-}]^y
ConditionResult
Q<KspQ \lt K_{sp}No precipitate (unsaturated)
Q=KspQ = K_{sp}Saturated, at equilibrium
Q>KspQ \gt K_{sp}Precipitate forms

Common Pitfalls

  • KspK_{sp} expressions do not include the concentration of the solid.
  • Solids and pure liquids are excluded from equilibrium expressions.
  • The common ion effect does not change KspK_{sp} itself — it shifts the equilibrium position.

Practice Problems

Problem 1

Calculate the pH\mathrm{pH} of a buffer prepared by mixing 100mL100\mathrm{ mL} of 0.20M0.20\mathrm{ M} CH3COOH\mathrm{CH}_3\mathrm{COOH} with 50mL50\mathrm{ mL} of 0.20M0.20\mathrm{ M} NaOH\mathrm{NaOH}. (pKa=4.76\mathrm{p}K_a = 4.76)

Solution:

Moles of CH3COOH=0.100×0.20=0.020mol\mathrm{CH}_3\mathrm{COOH} = 0.100 \times 0.20 = 0.020\mathrm{ mol}

Moles of NaOH=0.050×0.20=0.010mol\mathrm{NaOH} = 0.050 \times 0.20 = 0.010\mathrm{ mol}

NaOH\mathrm{NaOH} neutralises half the acid:

N(CH3COOH)remaining=0.020−0.010=0.010molN(\mathrm{CH}_3\mathrm{COOH})_{\mathrm{remaining}} = 0.020 - 0.010 = 0.010\mathrm{ mol}N(CH3COO−)formed=0.010molN(\mathrm{CH}_3\mathrm{COO}^-)_{\mathrm{formed}} = 0.010\mathrm{ mol}

Total volume =150mL=0.150L= 150\mathrm{ mL} = 0.150\mathrm{ L}:

[CH3COOH]=0.0100.150=0.0667M[\mathrm{CH}_3\mathrm{COOH}] = \frac{0.010}{0.150} = 0.0667\mathrm{ M}[CH3COO−]=0.0100.150=0.0667M[\mathrm{CH}_3\mathrm{COO}^-] = \frac{0.010}{0.150} = 0.0667\mathrm{ M}pH=4.76+log⁡0.06670.0667=4.76+0=4.76\mathrm{pH} = 4.76 + \log\frac{0.0667}{0.0667} = 4.76 + 0 = 4.76
Problem 2

Will a precipitate form when 50.0mL50.0\mathrm{ mL} of 1.0×10−4M1.0 \times 10^{-4}\mathrm{ M} AgNO3\mathrm{AgNO}_3 Is mixed with 50.0mL50.0\mathrm{ mL} of 1.0×10−4M1.0 \times 10^{-4}\mathrm{ M} NaCl\mathrm{NaCl}? (Ksp(AgCl)=1.8×10−10K_{sp}(\mathrm{AgCl}) = 1.8 \times 10^{-10})

Solution:

Total volume =100mL= 100\mathrm{ mL}. Diluted concentrations:

[Ag+]=0.050×1.0×10−40.100=5.0×10−5M[\mathrm{Ag}^+] = \frac{0.050 \times 1.0 \times 10^{-4}}{0.100} = 5.0 \times 10^{-5}\mathrm{ M}[Cl−]=0.050×1.0×10−40.100=5.0×10−5M[\mathrm{Cl}^-] = \frac{0.050 \times 1.0 \times 10^{-4}}{0.100} = 5.0 \times 10^{-5}\mathrm{ M}Q=(5.0×10−5)2=2.5×10−9Q = (5.0 \times 10^{-5})^2 = 2.5 \times 10^{-9}

Since Q=2.5×10−9>Ksp=1.8×10−10Q = 2.5 \times 10^{-9} \gt K_{sp} = 1.8 \times 10^{-10}A precipitate of AgCl\mathrm{AgCl} Will form.

Problem 3

Calculate the pH\mathrm{pH} of 0.050M0.050\mathrm{ M} NH3\mathrm{NH}_3 (Kb=1.8×10−5K_b = 1.8 \times 10^{-5}).

Solution:

NH3+H2O⇌NH4++OH−\mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{NH}_4^+ + \mathrm{OH}^-Kb=[NH4+][OH−][NH3]=x20.050−xK_b = \frac{[\mathrm{NH}_4^+][\mathrm{OH}^-]}{[\mathrm{NH}_3]} = \frac{x^2}{0.050 - x}

Approximation: x=1.8×10−5×0.050=9.0×10−7=9.5×10−4x = \sqrt{1.8 \times 10^{-5} \times 0.050} = \sqrt{9.0 \times 10^{-7}} = 9.5 \times 10^{-4}

Check: 9.5×10−4/0.050=1.9%<5%9.5 \times 10^{-4} / 0.050 = 1.9\% \lt 5\%. Valid.

pOH=−log⁡(9.5×10−4)=3.02\mathrm{pOH} = -\log(9.5 \times 10^{-4}) = 3.02pH=14.00−3.02=10.98\mathrm{pH} = 14.00 - 3.02 = 10.98
Problem 4

Explain why adding NH4Cl\mathrm{NH}_4\mathrm{Cl} to an NH3\mathrm{NH}_3 solution decreases the pH\mathrm{pH}.

Solution:

NH4Cl\mathrm{NH}_4\mathrm{Cl} dissociates completely to give NH4+\mathrm{NH}_4^+The conjugate acid of NH3\mathrm{NH}_3. This is the common ion effect:

NH3+H2O⇌NH4++OH−\mathrm{NH}_3 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{NH}_4^+ + \mathrm{OH}^-

Adding NH4+\mathrm{NH}_4^+ shifts the equilibrium to the left (Le Chatelier’s principle), decreasing [OH−][\mathrm{OH}^-] and therefore increasing [H+][\mathrm{H}^+]Which lowers the pH\mathrm{pH}.

Worked Examples

Worked Example: pH of a very dilute strong acid

Calculate the pH\mathrm{pH} of 1.0×10−8M1.0 \times 10^{-8}\mathrm{ M} HCl\mathrm{HCl} at 25°C25\degree\mathrm{C}.

Solution

At this concentration, the contribution from water autoionization ([H+]=10−7M[\mathrm{H}^+] = 10^{-7}\mathrm{ M}) is Significant and cannot be ignored. Let x=[OH−]x = [\mathrm{OH}^-] from water autoionization:

[H+]total=1.0×10−8+x[\mathrm{H}^+]_{\mathrm{total}} = 1.0 \times 10^{-8} + x

Kw=[H+][OH−]=(1.0×10−8+x)(x)=1.0×10−14K_w = [\mathrm{H}^+][\mathrm{OH}^-] = (1.0 \times 10^{-8} + x)(x) = 1.0 \times 10^{-14}

x2+1.0×10−8x−1.0×10−14=0x^2 + 1.0 \times 10^{-8}x - 1.0 \times 10^{-14} = 0

Using the quadratic formula:

x=−1.0×10−8+(1.0×10−8)2+4(1.0×10−14)2=−1.0×10−8+2.00×10−72=9.5×10−8x = \frac{-1.0 \times 10^{-8} + \sqrt{(1.0 \times 10^{-8})^2 + 4(1.0 \times 10^{-14})}}{2} = \frac{-1.0 \times 10^{-8} + 2.00 \times 10^{-7}}{2} = 9.5 \times 10^{-8}

[H+]total=1.0×10−8+9.5×10−8=1.05×10−7M[\mathrm{H}^+]_{\mathrm{total}} = 1.0 \times 10^{-8} + 9.5 \times 10^{-8} = 1.05 \times 10^{-7}\mathrm{ M}

pH=−log⁡(1.05×10−7)=6.98\mathrm{pH} = -\log(1.05 \times 10^{-7}) = 6.98

The pH\mathrm{pH} is close to 7 but slightly acidic, as expected for a very dilute strong acid. Ignoring Water autoionization would give the incorrect result pH=8.00\mathrm{pH} = 8.00 (a basic pH\mathrm{pH} from Adding acid), which violates chemical intuition.

Worked Example: pH at the equivalence point of a weak acid—strong base titration

Calculate the pH\mathrm{pH} at the equivalence point when 25.0mL25.0\mathrm{ mL} of 0.100M0.100\mathrm{ M} CH3COOH\mathrm{CH}_3\mathrm{COOH} (Ka=1.8×10−5K_a = 1.8 \times 10^{-5}) is titrated with 0.100M0.100\mathrm{ M} NaOH\mathrm{NaOH}.

Solution

At the equivalence point, all of the weak acid has been converted to its conjugate base:

n(CH3COOH)=0.0250×0.100=0.00250moln(\mathrm{CH}_3\mathrm{COOH}) = 0.0250 \times 0.100 = 0.00250\mathrm{ mol}

V(NaOH)=0.002500.100=25.0mLV(\mathrm{NaOH}) = \frac{0.00250}{0.100} = 25.0\mathrm{ mL}

Vtotal=50.0mL=0.0500LV_{\mathrm{total}} = 50.0\mathrm{ mL} = 0.0500\mathrm{ L}

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

The conjugate base hydrolyses water:

CH3COO−+H2O⇌CH3COOH+OH−\mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{CH}_3\mathrm{COOH} + \mathrm{OH}^-

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=5.27×10−6M[\mathrm{OH}^-] = \sqrt{K_b \times [\mathrm{CH}_3\mathrm{COO}^-]} = \sqrt{5.56 \times 10^{-10} \times 0.0500} = 5.27 \times 10^{-6}\mathrm{ M}

pOH=−log⁡(5.27×10−6)=5.28\mathrm{pOH} = -\log(5.27 \times 10^{-6}) = 5.28

pH=14.00−5.28=8.72\mathrm{pH} = 14.00 - 5.28 = 8.72

The equivalence point is basic because the conjugate base of a weak acid is itself a weak base. Phenolphthalein (transition range 8.3—10.0) is a suitable indicator. Bromothymol blue (6.0—7.6) would change colour Before reaching the equivalence point and would give a systematically low titre reading.

Worked Example: Selective precipitation using KspK_{sp}

A solution contains 0.020M0.020\mathrm{ M} Cl−\mathrm{Cl}^- and 0.020M0.020\mathrm{ M} CrO42−\mathrm{CrO}_4^{2-}. Solid AgNO3\mathrm{AgNO}_3 is added gradually. Ksp(AgCl)=1.8×10−10K_{sp}(\mathrm{AgCl}) = 1.8 \times 10^{-10} Ksp(Ag2CrO4)=1.1×10−12K_{sp}(\mathrm{Ag}_2\mathrm{CrO}_4) = 1.1 \times 10^{-12}. Determine which salt precipitates first, the [Ag+][\mathrm{Ag}^+] at which precipitation begins, and the [Ag+][\mathrm{Ag}^+] at which the second salt begins To precipitate.

Solution

Step 1: Calculate the threshold [Ag+][\mathrm{Ag}^+] for each salt.

For AgCl\mathrm{AgCl}:

[Ag+]=Ksp[Cl−]=1.8×10−100.020=9.0×10−9M[\mathrm{Ag}^+] = \frac{K_{sp}}{[\mathrm{Cl}^-]} = \frac{1.8 \times 10^{-10}}{0.020} = 9.0 \times 10^{-9}\mathrm{ M}

For Ag2CrO4\mathrm{Ag}_2\mathrm{CrO}_4:

[Ag+]=Ksp[CrO42−]=1.1×10−120.020=7.4×10−6M[\mathrm{Ag}^+] = \sqrt{\frac{K_{sp}}{[\mathrm{CrO}_4^{2-}]}} = \sqrt{\frac{1.1 \times 10^{-12}}{0.020}} = 7.4 \times 10^{-6}\mathrm{ M}

Step 2: Identify which precipitates first.

AgCl\mathrm{AgCl} precipitates first since it requires a lower [Ag+][\mathrm{Ag}^+] (9.0×10−9M9.0 \times 10^{-9}\mathrm{ M} Vs 7.4×10−6M7.4 \times 10^{-6}\mathrm{ M}).

Step 3: Calculate [Cl−][\mathrm{Cl}^-] remaining when Ag2CrO4\mathrm{Ag}_2\mathrm{CrO}_4 begins to precipitate.

When [Ag+]=7.4×10−6M[\mathrm{Ag}^+] = 7.4 \times 10^{-6}\mathrm{ M}:

[Cl−]remaining=Ksp[Ag+]=1.8×10−107.4×10−6=2.4×10−5M[\mathrm{Cl}^-]_{\mathrm{remaining}} = \frac{K_{sp}}{[\mathrm{Ag}^+]} = \frac{1.8 \times 10^{-10}}{7.4 \times 10^{-6}} = 2.4 \times 10^{-5}\mathrm{ M}

Fraction of Cl−\mathrm{Cl}^- precipitated: 0.020−2.4×10−50.020×100%=99.9%\dfrac{0.020 - 2.4 \times 10^{-5}}{0.020} \times 100\% = 99.9\%

This analysis underpins Mohr’s method for argentometric determination of chloride: the red colour of Ag2CrO4\mathrm{Ag}_2\mathrm{CrO}_4 appears only after virtually all Cl−\mathrm{Cl}^- has been removed.

Worked Example: Buffer preparation from a weak base

Prepare an NH3\mathrm{NH}_3/NH4Cl\mathrm{NH}_4\mathrm{Cl} buffer with pH=9.50\mathrm{pH} = 9.50. Given Kb(NH3)=1.8×10−5K_b(\mathrm{NH}_3) = 1.8 \times 10^{-5}. Calculate the mass of NH4Cl\mathrm{NH}_4\mathrm{Cl} (M=53.49g/molM = 53.49\mathrm{ g/mol}) that must be dissolved in 500mL500\mathrm{ mL} of 0.200M0.200\mathrm{ M} NH3\mathrm{NH}_3.

Solution

Step 1: Find the pKa\mathrm{p}K_a of the conjugate acid NH4+\mathrm{NH}_4^+.

pKb=−log⁡(1.8×10−5)=4.74\mathrm{p}K_b = -\log(1.8 \times 10^{-5}) = 4.74

pKa=14.00−4.74=9.26\mathrm{p}K_a = 14.00 - 4.74 = 9.26

Step 2: Apply the Henderson-Hasselbalch equation.

9.50=9.26+log⁡[NH3][NH4+]9.50 = 9.26 + \log\frac{[\mathrm{NH}_3]}{[\mathrm{NH}_4^+]}

log⁡[NH3][NH4+]=0.24\log\frac{[\mathrm{NH}_3]}{[\mathrm{NH}_4^+]} = 0.24

[NH3][NH4+]=100.24=1.74\frac{[\mathrm{NH}_3]}{[\mathrm{NH}_4^+]} = 10^{0.24} = 1.74

Step 3: Solve for the required [NH4+][\mathrm{NH}_4^+].

[NH4+]=0.2001.74=0.115M[\mathrm{NH}_4^+] = \frac{0.200}{1.74} = 0.115\mathrm{ M}

Step 4: Calculate the mass of NH4Cl\mathrm{NH}_4\mathrm{Cl}.

n(NH4Cl)=0.115×0.500=0.0575moln(\mathrm{NH}_4\mathrm{Cl}) = 0.115 \times 0.500 = 0.0575\mathrm{ mol}

m(NH4Cl)=0.0575×53.49=3.08gm(\mathrm{NH}_4\mathrm{Cl}) = 0.0575 \times 53.49 = 3.08\mathrm{ g}

Worked Example: Titration curve analysis for a polyprotic acid

25.0mL25.0\mathrm{ mL} of 0.100M0.100\mathrm{ M} H3PO4\mathrm{H}_3\mathrm{PO}_4 (Ka1=7.5×10−3K_{a1} = 7.5 \times 10^{-3} Ka2=6.2×10−8K_{a2} = 6.2 \times 10^{-8}, Ka3=4.8×10−13K_{a3} = 4.8 \times 10^{-13}) is titrated with 0.100M0.100\mathrm{ M} NaOH\mathrm{NaOH}. Calculate the pH\mathrm{pH} at the first and second equivalence points and identify Suitable indicators.

Solution

First equivalence point (25.0mL25.0\mathrm{ mL} NaOH\mathrm{NaOH} added):

All H3PO4\mathrm{H}_3\mathrm{PO}_4 is converted to H2PO4−\mathrm{H}_2\mathrm{PO}_4^- (an amphoteric species).

[H2PO4−]=0.002500.0500=0.0500M[\mathrm{H}_2\mathrm{PO}_4^-] = \frac{0.00250}{0.0500} = 0.0500\mathrm{ M}

For an amphoteric species, the pH\mathrm{pH} is approximately the average of the two relevant pKa\mathrm{p}K_a values:

pH≈pKa1+pKa22=2.12+7.212=4.67\mathrm{pH} \approx \frac{\mathrm{p}K_{a1} + \mathrm{p}K_{a2}}{2} = \frac{2.12 + 7.21}{2} = 4.67

A suitable indicator: bromocresol green (3.8—5.4) or methyl red (4.4—6.2).

Second equivalence point (50.0mL50.0\mathrm{ mL} NaOH\mathrm{NaOH} added):

All H2PO4−\mathrm{H}_2\mathrm{PO}_4^- is converted to HPO42−\mathrm{HPO}_4^{2-} (also amphoteric).

[HPO42−]=0.002500.0750=0.0333M[\mathrm{HPO}_4^{2-}] = \frac{0.00250}{0.0750} = 0.0333\mathrm{ M}

pH≈pKa2+pKa32=7.21+12.322=9.76\mathrm{pH} \approx \frac{\mathrm{p}K_{a2} + \mathrm{p}K_{a3}}{2} = \frac{7.21 + 12.32}{2} = 9.76

A suitable indicator: phenolphthalein (8.3—10.0).

Third equivalence point: Not achievable in aqueous solution because Ka3K_{a3} is too small (4.8×10−134.8 \times 10^{-13}) for complete neutralisation of the third proton.

Common Pitfalls

  • Assuming complete dissociation for all diprotic acids: H2SO4\mathrm{H}_2\mathrm{SO}_4 has a complete first dissociation but a partial second (Ka2=1.2×10−2K_{a2} = 1.2 \times 10^{-2}), so [H+]≠2[acid][\mathrm{H}^+] \neq 2[\mathrm{acid}]. For H2CO3\mathrm{H}_2\mathrm{CO}_3Both dissociation steps are weak. Always check the magnitude of each KaK_a before making simplifying assumptions.

  • Using pOH=14−pH\mathrm{pOH} = 14 - \mathrm{pH} without specifying temperature: Kw=1.0×10−14K_w = 1.0 \times 10^{-14} only at 25°C25\degree\mathrm{C}. At 37°C37\degree\mathrm{C}, Kw≈2.4×10−14K_w \approx 2.4 \times 10^{-14}So pH+pOH=13.62\mathrm{pH} + \mathrm{pOH} = 13.62. Always state the temperature assumption explicitly.

  • Applying Henderson-Hasselbalch to strong acid—base mixtures: The equation requires a weak acid and its conjugate base. For a strong acid titrated with a strong base, calculate the excess [H+][\mathrm{H}^+] or [OH−][\mathrm{OH}^-] directly from stoichiometry.

  • Forgetting dilution when mixing buffer components: When two solutions are combined to make a buffer, the total volume changes. Convert all quantities to moles first, then recalculate concentrations in the combined volume before applying the Henderson-Hasselbalch equation.

  • Comparing KspK_{sp} values across different stoichiometries: Ksp(AgCl)=1.8×10−10K_{sp}(\mathrm{AgCl}) = 1.8 \times 10^{-10} and Ksp(Ag2CrO4)=1.1×10−12K_{sp}(\mathrm{Ag}_2\mathrm{CrO}_4) = 1.1 \times 10^{-12}But Ag2CrO4\mathrm{Ag}_2\mathrm{CrO}_4 is more soluble in water because its KspK_{sp} expression contains [Ag+]2[\mathrm{Ag}^+]^2. Always calculate molar solubility from KspK_{sp} before comparing.

  • Ignoring water autoionization for dilute solutions: When the calculated [H+][\mathrm{H}^+] from the acid or base alone is below 10−6M10^{-6}\mathrm{ M}The contribution from water (10−7M10^{-7}\mathrm{ M}) is comparable and must be included via the full quadratic.

  • Confusing buffer capacity with buffer range: Buffer capacity (total moles of acid or base that can be absorbed) depends on the absolute concentrations of the buffer components, not their ratio. A 0.01M0.01\mathrm{ M} buffer at pH=pKa\mathrm{pH} = \mathrm{p}K_a has far less capacity than a 1.0M1.0\mathrm{ M} buffer at the same pH\mathrm{pH}.

  • Assuming the 5% rule is always valid: The approximation [HA]≈c0[\mathrm{HA}] \approx c_0 fails when KaK_a is large relative to c0c_0. Always verify by computing x/c0×100%x/c_0 \times 100\%. If it exceeds 5%, solve the full quadratic.

  • Using the wrong pKa\mathrm{p}K_a in Henderson-Hasselbalch: When working with a weak base (e.g., NH3\mathrm{NH}_3), use the pKa\mathrm{p}K_a of its conjugate acid (NH4+\mathrm{NH}_4^+), not the pKb\mathrm{p}K_b of the base itself. The relationship is pKa+pKb=14.00\mathrm{p}K_a + \mathrm{p}K_b = 14.00.

  • Assuming the common ion effect changes KspK_{sp}: Adding a common ion shifts the equilibrium position (decreasing solubility), but KspK_{sp} itself is a thermodynamic constant that depends only on temperature.

Exam-Style Problems

  1. Calculate the pH\mathrm{pH} of the solution formed when 15.0mL15.0\mathrm{ mL} of 0.100M0.100\mathrm{ M} H2SO4\mathrm{H}_2\mathrm{SO}_4 is added to 35.0mL35.0\mathrm{ mL} of 0.100M0.100\mathrm{ M} NaOH\mathrm{NaOH} at 25°C25\degree\mathrm{C}. (Ka2K_{a2} of HSO4−\mathrm{HSO}_4^- = 1.2×10−21.2 \times 10^{-2}.) State all assumptions and justify their validity. [Medium]

  2. A buffer is prepared by dissolving 4.10g4.10\mathrm{ g} of sodium ethanoate (CH3COONa\mathrm{CH}_3\mathrm{COONa}, M=82.03g/molM = 82.03\mathrm{ g/mol}) in 250mL250\mathrm{ mL} of 0.200M0.200\mathrm{ M} ethanoic acid (pKa=4.76\mathrm{p}K_a = 4.76). (a) Calculate the buffer pH\mathrm{pH}. (b) Calculate the new pH\mathrm{pH} after adding 5.0mL5.0\mathrm{ mL} of 0.100M0.100\mathrm{ M} HCl\mathrm{HCl}. (c) Calculate the percentage change in pH\mathrm{pH} and comment on the effectiveness of the buffer. [Medium]

  3. Will a precipitate form when 25.0mL25.0\mathrm{ mL} of 2.0×10−4M2.0 \times 10^{-4}\mathrm{ M} Pb(NO3)2\mathrm{Pb}(\mathrm{NO}_3)_2 is mixed with 25.0mL25.0\mathrm{ mL} of 1.0×10−3M1.0 \times 10^{-3}\mathrm{ M} NaI\mathrm{NaI}? Ksp(PbI2)=7.1×10−9K_{sp}(\mathrm{PbI}_2) = 7.1 \times 10^{-9}. If a precipitate forms, calculate [Pb2+][\mathrm{Pb}^{2+}] and [I−][\mathrm{I}^-] remaining at equilibrium. [Hard]

  4. 20.0mL20.0\mathrm{ mL} of 0.150M0.150\mathrm{ M} NH3\mathrm{NH}_3 (Kb=1.8×10−5K_b = 1.8 \times 10^{-5}) is titrated with 0.150M0.150\mathrm{ M} HCl\mathrm{HCl}. Calculate the pH\mathrm{pH} at each of the following volumes of HCl\mathrm{HCl} added: (a) 0.0mL0.0\mathrm{ mL}(b) 10.0mL10.0\mathrm{ mL}(c) 20.0mL20.0\mathrm{ mL} (equivalence point), (d) 25.0mL25.0\mathrm{ mL}. Sketch the approximate titration curve and label the buffer region. [Hard]

  5. The solubility of CaF2\mathrm{CaF}_2 in pure water is 2.15×10−4M2.15 \times 10^{-4}\mathrm{ M} at 25°C25\degree\mathrm{C}. (a) Calculate KspK_{sp}. (b) Calculate the solubility of CaF2\mathrm{CaF}_2 in 0.050M0.050\mathrm{ M} CaCl2\mathrm{CaCl}_2. (c) Determine the maximum concentration of NaF\mathrm{NaF} that can coexist with 0.010M0.010\mathrm{ M} CaCl2\mathrm{CaCl}_2 without precipitation. [Medium]

  6. Explain why phenolphthalein is a suitable indicator for the titration of ethanoic acid with sodium hydroxide, but methyl orange is not. Support your answer with a quantitative calculation of the equivalence point pH\mathrm{pH} and reference to the transition ranges of both indicators. [Medium]

  7. A student prepares a buffer by mixing 100mL100\mathrm{ mL} of 0.200M0.200\mathrm{ M} CH3COOH\mathrm{CH}_3\mathrm{COOH} with 100mL100\mathrm{ mL} of 0.100M0.100\mathrm{ M} NaOH\mathrm{NaOH}. (a) Calculate the buffer pH\mathrm{pH}. (b) Determine the maximum volume of 0.100M0.100\mathrm{ M} HCl\mathrm{HCl} that can be added before the pH\mathrm{pH} drops below 4.004.00. (c) Comment on whether this buffer would be effective at pH=4.00\mathrm{pH} = 4.00 given the pKa\mathrm{p}K_a of ethanoic acid. [Hard]

  8. An environmental scientist measures the pH\mathrm{pH} of a lake at 4.504.50. (a) Calculate [H+][\mathrm{H}^+] and [SO42−][\mathrm{SO}_4^{2-}]Assuming the acidity is entirely from dissolved H2SO4\mathrm{H}_2\mathrm{SO}_4 with complete dissociation of both protons. (b) Determine whether CaSO4\mathrm{CaSO}_4 would precipitate if [Ca2+]=1.5×10−3M[\mathrm{Ca}^{2+}] = 1.5 \times 10^{-3}\mathrm{ M}. Ksp(CaSO4)=2.4×10−5K_{sp}(\mathrm{CaSO}_4) = 2.4 \times 10^{-5}. (c) Calculate the minimum [Ca2+][\mathrm{Ca}^{2+}] required to initiate precipitation of CaSO4\mathrm{CaSO}_4. [Medium]

If You Get These Wrong, Revise:

Summary

This topic covers the essential chemistry of acids and bases (advanced), 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.