Chemistry - Acids, Bases, and Electrochemistry

Acids and Bases

Definitions

Arrhenius Theory:

• Acid: produces $\mathrm{H}^+$ (or $\mathrm{H}_3\mathrm{O}^+$) in aqueous solution
• Base: produces $\mathrm{OH}^-$ in aqueous solution

Bronsted-Lowry Theory:

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

Lewis Theory:

• Acid: electron pair acceptor
• Base: electron pair donor

:::info In the DSE syllabus, the Bronsted-Lowry definition is most commonly used. Always identify The proton donor and proton acceptor in acid-base reactions. :::

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.

$\mathrm{HA} + \mathrm{B} \rightleftharpoons \mathrm{A}^- + \mathrm{BH}^+$

• $\mathrm{HA}$ and $\mathrm{A}^-$ form a conjugate acid-base pair
• $\mathrm{B}$ and $\mathrm{BH}^+$ form a conjugate acid-base pair

Strong and Weak Acids

:::caution Warning Ionisation; concentration refers to the amount dissolved per unit volume. A dilute solution of a Strong acid can have a higher pH than a concentrated solution of a weak acid. :::

Common Strong Acids and Bases

Strong acids: \mathrm{HCl}$$\mathrm{HBr}$$\mathrm{HI}$$\mathrm{HNO}_3 \mathrm{H}_2\mathrm{SO}_4$$\mathrm{HClO}_4

Strong bases: Group 1 hydroxides (\mathrm{NaOH}$$\mathrm{KOH}), $\mathrm{Ba(OH)}_2$

The pH Scale

Definition of pH

$\mathrm{pH} = -\log_{10}[\mathrm{H}^+]$

Where $[\mathrm{H}^+]$ is the concentration of hydrogen ions in mol/dm$^3$.

pH of Water

Pure water at $25^\circ\mathrm{C}$: $[\mathrm{H}^+] = [\mathrm{OH}^-] = 10^{-7}$ mol/dm$^3$So $\mathrm{pH} = 7$.

The ionic product of water:

$K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 10^{-14} \mathrm{ at } 25^\circ\mathrm{C}$

This relationship always holds for aqueous solutions at $25^\circ\mathrm{C}$.

Worked Example 1

Find the pH of a $0.05 \mathrm{ mol/dm}^3$ solution of $\mathrm{HCl}$.

$\mathrm{HCl}$ is a strong acid, so it is fully ionised:

$[\mathrm{H}^+] = 0.05 \mathrm{ mol/dm}^3$

$\mathrm{pH} = -\log_{10}(0.05) = -\log_{10}(5 \times 10^{-2}) = 2 - \log_{10}5 = 2 - 0.699 = 1.30$

Worked Example 2

Find the pH of a $0.1 \mathrm{ mol/dm}^3$ solution of $\mathrm{NaOH}$.

$\mathrm{NaOH}$ is a strong base, fully ionised:

$[\mathrm{OH}^-] = 0.1 \mathrm{ mol/dm}^3$

$[\mathrm{H}^+] = \frac{K_w}{[\mathrm{OH}^-]} = \frac{10^{-14}}{0.1} = 10^{-13} \mathrm{ mol/dm}^3$

$\mathrm{pH} = -\log_{10}(10^{-13}) = 13$

Worked Example 3

Find the pH of a $0.1 \mathrm{ mol/dm}^3$ solution of $\mathrm{CH}_3\mathrm{COOH}$ ($K_a = 1.8 \times 10^{-5}$).

For a weak acid:

$K_a = \frac{[\mathrm{H}^+][\mathrm{CH}_3\mathrm{COO}^-]}{[\mathrm{CH}_3\mathrm{COOH}]}$

Assuming $[\mathrm{H}^+] = x$:

$1.8 \times 10^{-5} = \frac{x^2}{0.1}$

$x^2 = 1.8 \times 10^{-6}$

$x = 1.34 \times 10^{-3} \mathrm{ mol/dm}^3$

$\mathrm{pH} = -\log_{10}(1.34 \times 10^{-3}) = 2.87$

:::tip Tip $[\mathrm{HA}]_{\mathrm{initial}} \approx [\mathrm{HA}]_{\mathrm{equilibrium}}$ only when the degree Of ionisation is small ( when $K_a \lt 10^{-4}$). This simplification is valid for most DSE-level problems. :::

Worked Example: pH After Dilution

A solution of $\mathrm{HCl}$ has $\mathrm{pH} = 2.00$. If $10.0 \mathrm{ cm^3}$ of this solution is diluted to $250 \mathrm{ cm^3}$What is the new pH?

Original $[\mathrm{H^+}] = 10^{-2.00} = 0.0100 \mathrm{ mol/dm^3}$

After dilution: $[\mathrm{H^+}] = 0.0100 \times \frac{10.0}{250} = 4.00 \times 10^{-4} \mathrm{ mol/dm^3}$

$\mathrm{pH} = -\log_{10}(4.00 \times 10^{-4}) = 3.40$

Worked Example: Identifying Conjugate Pairs

In the reaction $\mathrm{HNO_2} + \mathrm{H_2O} \rightleftharpoons \mathrm{NO_2^-} + \mathrm{H_3O^+}$Identify the two conjugate acid-base pairs.

$\mathrm{HNO_2}$ donates a proton to become $\mathrm{NO_2^-}$:

• Conjugate pair 1: $\mathrm{HNO_2}$ (acid) / $\mathrm{NO_2^-}$ (conjugate base)

$\mathrm{H_2O}$ accepts a proton to become $\mathrm{H_3O^+}$:

• Conjugate pair 2: $\mathrm{H_2O}$ (base) / $\mathrm{H_3O^+}$ (conjugate acid)

pH Scale Summary

Each unit change in pH represents a tenfold change in $[\mathrm{H}^+]$.

Acid-Base Titrations

A titration is a technique for determining the concentration of a solution by reacting it with a Solution of known concentration.

Titration Procedure

1. Rinse the burette with the solution it will contain, then fill it
2. Record the initial burette reading
3. Add indicator to the solution in the conical flask
4. Slowly add the titrant from the burette, swirling constantly
5. Stop when the indicator changes colour (endpoint)
6. Record the final burette reading
7. Repeat until concordant results are obtained (within $0.10 \mathrm{ cm}^3$)

Indicators

Worked Example 4

$25.0 \mathrm{ cm}^3$ of $\mathrm{NaOH}$ solution is titrated with $0.100 \mathrm{ mol/dm}^3$ $\mathrm{HCl}$. The average titre is $20.0 \mathrm{ cm}^3$. Find the concentration of $\mathrm{NaOH}$.

$\mathrm{HCl} + \mathrm{NaOH} \to \mathrm{NaCl} + \mathrm{H}_2\mathrm{O}$

Moles of $\mathrm{HCl} = 0.100 \times \frac{20.0}{1000} = 0.00200 \mathrm{ mol}$

Since the mole ratio is 1:1:

Moles of $\mathrm{NaOH} = 0.00200 \mathrm{ mol}$

$[\mathrm{NaOH}] = \frac{0.00200}{25.0/1000} = \frac{0.00200}{0.0250} = 0.0800 \mathrm{ mol/dm}^3$

Salt Preparation

Solubility Rules

Methods of Salt Preparation

Soluble salt from acid + insoluble base:

1. Add excess insoluble base (metal oxide or carbonate) to the acid
2. Filter to remove excess base
3. Evaporate the filtrate to crystallisation
4. Filter and dry the crystals

Soluble salt by titration (acid + soluble base):

1. Titrate to find the exact volumes needed
2. Repeat using exact volumes without indicator
3. Evaporate to crystallisation

Insoluble salt by precipitation:

1. Mix two soluble salts that contain the required ions
2. Filter the precipitate
3. Wash with distilled water
4. Dry between filter papers

Worked Example 5

Describe how to prepare a sample of copper(II) sulphate crystals.

1. Add dilute $\mathrm{H}_2\mathrm{SO}_4$ to a beaker
2. Add copper(II) oxide powder in excess (it is an insoluble base)
3. Warm gently and stir until no more reacts
4. Filter to remove excess copper(II) oxide
5. Evaporate the filtrate until crystals start to form
6. Leave to cool and crystallise
7. Filter, wash with cold water, and dry

Redox Reactions

Oxidation and Reduction

Oxidation: Loss of electrons, increase in oxidation number

Reduction: Gain of electrons, decrease in oxidation number

OIL RIG: Oxidation Is Loss, Reduction Is Gain

LEO says GER: Losing Electrons is Oxidation, Gaining Electrons is Reduction

Oxidation Numbers

Rules for assigning oxidation numbers:

1. Elements in their standard state have oxidation number 0 (e.g., \mathrm{Na}$$\mathrm{Cl}_2 $\mathrm{O}_2$)
2. Simple ions have oxidation number equal to their charge (e.g., $\mathrm{Na}^+ = +1$ $\mathrm{Cl}^- = -1$)
3. Oxygen is -2 (except in peroxides: -1; in $\mathrm{OF}_2$: +2)
4. Hydrogen is +1 (except in metal hydrides: -1)
5. The sum of oxidation numbers in a neutral compound is 0
6. The sum of oxidation numbers in a polyatomic ion equals the charge on the ion

Worked Example 6

Find the oxidation numbers of each element in $\mathrm{KMnO}_4$.

Let the oxidation number of Mn be $x$.

$+1 + x + 4(-2) = 0$

$1 + x - 8 = 0$

$x = +7$

Oxidation numbers: \mathrm{K} = +1$$\mathrm{Mn} = +7$$\mathrm{O} = -2.

Balancing Redox Equations (Ion-Electron Method)

1. Split the equation into two half-equations (oxidation and reduction)
2. Balance atoms other than $\mathrm{O}$ and $\mathrm{H}$
3. Balance $\mathrm{O}$ by adding $\mathrm{H}_2\mathrm{O}$
4. Balance $\mathrm{H}$ by adding $\mathrm{H}^+$
5. Balance charge by adding electrons ($e^-$)
6. Multiply half-equations so that the electrons cancel
7. Add the half-equations and simplify

Worked Example 7

Balance the reaction: $\mathrm{MnO}_4^- + \mathrm{Fe}^{2+} \to \mathrm{Mn}^{2+} + \mathrm{Fe}^{3+}$ (in acidic solution)

Reduction half-equation:

$\mathrm{MnO}_4^- \to \mathrm{Mn}^{2+}$

$\mathrm{MnO}_4^- + 8\mathrm{H}^+ \to \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$

Charge: $-1 + 8 = +7$ (left), $+2$ (right). Add $5e^-$ to left:

$\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \to \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$

Oxidation half-equation:

$\mathrm{Fe}^{2+} \to \mathrm{Fe}^{3+} + e^-$

Multiply by 5:

$5\mathrm{Fe}^{2+} \to 5\mathrm{Fe}^{3+} + 5e^-$

Combine:

$\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5\mathrm{Fe}^{2+} \to \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O} + 5\mathrm{Fe}^{3+}$

Electrolysis

Definitions

Electrolysis: The decomposition of an ionic compound by passing an electric current through it.

Electrolyte: The ionic compound, either molten or in aqueous solution, that conducts Electricity.

Electrodes: Conductors through which current enters and leaves the electrolyte.

• Anode (+): Positive electrode where oxidation occurs
• Cathode (-): Negative electrode where reduction occurs

Electrolysis of Molten Ionic Compounds

At the cathode (reduction): Metal ions gain electrons and are discharged as metal atoms.

$\mathrm{M}^{n+} + ne^- \to \mathrm{M}$

At the anode (oxidation): Non-metal ions lose electrons and are discharged.

$\mathrm{X}^{n-} \to \frac{n}{2}\mathrm{X}_2 + ne^-$

Worked Example 8

Describe the electrolysis of molten lead(II) bromide, $\mathrm{PbBr}_2$.

At the cathode (-): $\mathrm{Pb}^{2+} + 2e^- \to \mathrm{Pb}$ (grey solid)

At the anode (+): $2\mathrm{Br}^- \to \mathrm{Br}_2 + 2e^-$ (orange-brown gas)

Electrolysis of Aqueous Solutions

When an aqueous solution is electrolysed, both the dissolved ions and water molecules can be Discharged. The discharge series determines which species is preferentially discharged:

At the cathode (less reactive metal is discharged):

$\mathrm{K}^+ \lt \mathrm{Na}^+ \lt \mathrm{Ca}^{2+} \lt \mathrm{Mg}^{2+} \lt \mathrm{Al}^{3+} \lt \mathrm{Zn}^{2+} \lt \mathrm{Fe}^{2+} \lt \mathrm{Ni}^{2+} \lt \mathrm{Sn}^{2+} \lt \mathrm{Pb}^{2+} \lt \mathrm{H}^+ \lt \mathrm{Cu}^{2+} \lt \mathrm{Ag}^+ \lt \mathrm{Au}^+$

Ions above $\mathrm{H}^+$: $\mathrm{H}_2\mathrm{O}$ is reduced instead ($2\mathrm{H}_2\mathrm{O} + 2e^- \to \mathrm{H}_2 + 2\mathrm{OH}^-$)

Ions below $\mathrm{H}^+$: The metal ion is discharged

At the anode:

$\mathrm{SO}_4^{2-} \lt \mathrm{NO}_3^- \lt \mathrm{Cl}^- \lt \mathrm{Br}^- \lt \mathrm{I}^- \lt \mathrm{OH}^-$

Sulphate and nitrate: $\mathrm{H}_2\mathrm{O}$ is oxidised instead ($4\mathrm{OH}^- \to \mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O} + 4e^-$Or $2\mathrm{H}_2\mathrm{O} \to \mathrm{O}_2 + 4\mathrm{H}^+ + 4e^-$)

Halides (\mathrm{Cl}^-$$\mathrm{Br}^-$$\mathrm{I}^-): The halogen is discharged

:::caution Warning At the anode instead of $\mathrm{OH}^-$. In dilute chloride solutions, $\mathrm{OH}^-$ may be Preferentially discharged. :::

Worked Example 9

Describe the electrolysis of concentrated aqueous $\mathrm{NaCl}$ using carbon electrodes.

At the cathode: $\mathrm{Na}^+$ is above $\mathrm{H}^+$ in the discharge series, so $\mathrm{H}_2\mathrm{O}$ is reduced:

$2\mathrm{H}_2\mathrm{O} + 2e^- \to \mathrm{H}_2 + 2\mathrm{OH}^-$

At the anode: Concentrated $\mathrm{Cl}^-$ is discharged (halides above $\mathrm{OH}^-$ in Concentrated solution):

$2\mathrm{Cl}^- \to \mathrm{Cl}_2 + 2e^-$

Overall: $2\mathrm{H}_2\mathrm{O} + 2\mathrm{NaCl} \to \mathrm{H}_2 + \mathrm{Cl}_2 + 2\mathrm{NaOH}$

Faraday”s Laws of Electrolysis

First Law

The mass of substance liberated at an electrode is proportional to the quantity of charge passed.

$m = \frac{Q \times M}{nF}$

Where:

• $m$ = mass liberated (g)
• $Q$ = charge (C) = $I \times t$ (current in A $\times$ time in s)
• $M$ = molar mass (g/mol)
• $n$ = number of electrons transferred per ion
• $F$ = Faraday constant $= 96500 \mathrm{ C/mol}$

Second Law

When the same quantity of electricity is passed through different electrolytes, the masses of Different substances liberated are proportional to their equivalent masses ($M/n$).

Worked Example 10

What mass of copper is deposited when a current of $2.0 \mathrm{ A}$ is passed through $\mathrm{CuSO}_4$ solution for 30 minutes?

$Q = It = 2.0 \times 30 \times 60 = 3600 \mathrm{ C}$

$\mathrm{Cu}^{2+} + 2e^- \to \mathrm{Cu} \quad (n = 2)$

$m = \frac{Q \times M}{nF} = \frac{3600 \times 63.5}{2 \times 96500} = \frac{228600}{193000} = 1.18 \mathrm{ g}$

Worked Example 11

What volume of oxygen (at r.t.p.) is produced when a current of $3.0 \mathrm{ A}$ is passed through Dilute $\mathrm{H}_2\mathrm{SO}_4$ for 20 minutes?

$Q = 3.0 \times 20 \times 60 = 3600 \mathrm{ C}$

At the anode: $4\mathrm{OH}^- \to \mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O} + 4e^-$ ($n = 4$)

Moles of $\mathrm{O}_2 = \frac{Q}{nF} = \frac{3600}{4 \times 96500} = 0.00933 \mathrm{ mol}$

Volume at r.t.p. ($1 \mathrm{ mol} = 24.0 \mathrm{ dm}^3$):

$V = 0.00933 \times 24.0 = 0.224 \mathrm{ dm}^3 = 224 \mathrm{ cm}^3$

Electrochemical Cells

Voltaic (Galvanic) Cells

A voltaic cell converts chemical energy to electrical energy through a spontaneous redox reaction.

Structure of a Voltaic Cell

1. Two half-cells, each containing an electrode in contact with an electrolyte
2. A metal wire connecting the two electrodes (external circuit)
3. A salt bridge or porous barrier connecting the two electrolytes (internal circuit)

Salt bridge: Contains an inert electrolyte (e.g., $\mathrm{KNO}_3$) that allows ions to flow Without the solutions mixing directly.

Electrode Potentials

The standard electrode potential ($E^\circ$) is the potential difference between a half-cell and The standard hydrogen electrode (SHE) under standard conditions (298 K, 1 mol/dm$^3$1 atm).

The SHE is assigned $E^\circ = 0.00 \mathrm{ V}$.

Standard Cell Potential

$E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{cathode}} - E^\circ_{\mathrm{anode}}$

Where:

• The cathode has the more positive (less negative) $E^\circ$ value (reduction occurs)
• The anode has the less positive (more negative) $E^\circ$ value (oxidation occurs)

If $E^\circ_{\mathrm{cell}} \gt 0$The reaction is spontaneous.

Worked Example 12

A cell is constructed from a $\mathrm{Zn}^{2+}/\mathrm{Zn}$ half-cell ($E^\circ = -0.76 \mathrm{ V}$) and a $\mathrm{Cu}^{2+}/\mathrm{Cu}$ half-cell ($E^\circ = +0.34 \mathrm{ V}$). Find the cell potential and write the overall equation.

Copper has the more positive $E^\circ$So reduction occurs at the copper electrode (cathode).

$E^\circ_{\mathrm{cell}} = 0.34 - (-0.76) = 1.10 \mathrm{ V}$

Cathode (reduction): $\mathrm{Cu}^{2+} + 2e^- \to \mathrm{Cu}$

Anode (oxidation): $\mathrm{Zn} \to \mathrm{Zn}^{2+} + 2e^-$

Overall: $\mathrm{Zn} + \mathrm{Cu}^{2+} \to \mathrm{Zn}^{2+} + \mathrm{Cu}$

The Electrochemical Series

The electrochemical series ranks half-reactions by their standard electrode potentials:

More negative $E^\circ$: Metal is a stronger reducing agent (more oxidised).

More positive $E^\circ$: Ion is a stronger oxidising agent (more reduced).

Summary Table

Exam Tips

• In titration calculations, always convert volumes to dm$^3$ by dividing by 1000.
• For weak acid pH calculations, set up the $K_a$ expression and solve the quadratic (or use the approximation).
• When writing redox half-equations, always balance charge with electrons last.
• In electrolysis, identify the ions present and use the discharge series to determine the products.
• For Faraday’s law problems, remember to convert minutes to seconds.
• In electrochemical cell questions, the species with the more positive $E^\circ$ undergoes reduction (cathode).

Question 1: $25.0 \mathrm{ cm}^3$ of $0.200 \mathrm{ mol/dm}^3$ $\mathrm{H}_2\mathrm{SO}_4$ is Neutralised by $\mathrm{NaOH}$ solution. If $20.0 \mathrm{ cm}^3$ of $\mathrm{NaOH}$ is required, Find its concentration.

$\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{NaOH} \to \mathrm{Na}_2\mathrm{SO}_4 + 2\mathrm{H}_2\mathrm{O}$

Moles of $\mathrm{H}_2\mathrm{SO}_4 = 0.200 \times 0.0250 = 0.00500 \mathrm{ mol}$

Moles of $\mathrm{NaOH} = 2 \times 0.00500 = 0.0100 \mathrm{ mol}$

$[\mathrm{NaOH}] = \frac{0.0100}{0.0200} = 0.500 \mathrm{ mol/dm}^3$

Question 2: A current of $5.0 \mathrm{ A}$ is passed through molten $\mathrm{Al}_2\mathrm{O}_3$ For 2 hours. What mass of aluminium is produced?

$Q = 5.0 \times 2 \times 3600 = 36000 \mathrm{ C}$

$\mathrm{Al}^{3+} + 3e^- \to \mathrm{Al} \quad (n = 3)$

$m = \frac{36000 \times 27.0}{3 \times 96500} = \frac{972000}{289500} = 3.36 \mathrm{ g}$

Question 3: Assign oxidation numbers to all elements in $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$.

$2(+1) + 2x + 7(-2) = 0$

$2 + 2x - 14 = 0$

$x = +6$

Oxidation numbers: \mathrm{K} = +1$$\mathrm{Cr} = +6$$\mathrm{O} = -2.

Question 4: Balance the reaction between $\mathrm{MnO}_4^-$ and $\mathrm{C}_2\mathrm{O}_4^{2-}$ In acidic solution.

Reduction: $\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \to \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$

Oxidation: $\mathrm{C}_2\mathrm{O}_4^{2-} \to 2\mathrm{CO}_2 + 2e^-$

Multiply oxidation by 5 and reduction by 2:

$2\mathrm{MnO}_4^- + 16\mathrm{H}^+ + 10e^- \to 2\mathrm{Mn}^{2+} + 8\mathrm{H}_2\mathrm{O}$

$5\mathrm{C}_2\mathrm{O}_4^{2-} \to 10\mathrm{CO}_2 + 10e^-$

Overall: $2\mathrm{MnO}_4^- + 5\mathrm{C}_2\mathrm{O}_4^{2-} + 16\mathrm{H}^+ \to 2\mathrm{Mn}^{2+} + 10\mathrm{CO}_2 + 8\mathrm{H}_2\mathrm{O}$

Question 5: A cell is made from $\mathrm{Zn}^{2+}/\mathrm{Zn}$ ($E^\circ = -0.76 \mathrm{ V}$) And $\mathrm{Ag}^+/\mathrm{Ag}$ ($E^\circ = +0.80 \mathrm{ V}$). Write the overall equation and Calculate the cell potential.

Silver has the more positive $E^\circ$So it is the cathode.

$E^\circ_{\mathrm{cell}} = 0.80 - (-0.76) = 1.56 \mathrm{ V}$

Cathode: $\mathrm{Ag}^+ + e^- \to \mathrm{Ag}$

Anode: $\mathrm{Zn} \to \mathrm{Zn}^{2+} + 2e^-$

Multiply cathode by 2: $2\mathrm{Ag}^+ + 2e^- \to 2\mathrm{Ag}$

Overall: $\mathrm{Zn} + 2\mathrm{Ag}^+ \to \mathrm{Zn}^{2+} + 2\mathrm{Ag}$

Buffers

Definition

A buffer solution is one that resists changes in pH when small amounts of acid or base are added.

Composition

A buffer can be made from:

• A weak acid and its conjugate base (salt of the weak acid)
• A weak base and its conjugate acid (salt of the weak base)

How a Buffer Works

Acidic buffer (e.g., $\mathrm{CH}_3\mathrm{COOH}$ / $\mathrm{CH}_3\mathrm{COONa}$):

When acid ($\mathrm{H}^+$) is added: $\mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}^+ \to \mathrm{CH}_3\mathrm{COOH}$ The conjugate base Neutralises the added $\mathrm{H}^+$.

When base ($\mathrm{OH}^-$) is added: $\mathrm{CH}_3\mathrm{COOH} + \mathrm{OH}^- \to \mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}_2\mathrm{O}$ The weak acid neutralises the added $\mathrm{OH}^-$.

Henderson-Hasselbalch Equation

$\mathrm{pH} = \mathrm{p}K_a + \log_{10}\left(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\right)$

Where:

• $\mathrm{p}K_a = -\log_{10}K_a$
• $[\mathrm{A}^-]$ = concentration of the conjugate base
• $[\mathrm{HA}]$ = concentration of the weak acid

Worked Example 12

A buffer contains $0.1 \mathrm{ mol/dm}^3$ $\mathrm{CH}_3\mathrm{COOH}$ ($K_a = 1.8 \times 10^{-5}$) And $0.2 \mathrm{ mol/dm}^3$ $\mathrm{CH}_3\mathrm{COONa}$. Calculate the pH.

$\mathrm{p}K_a = -\log_{10}(1.8 \times 10^{-5}) = 4.74$

$\mathrm{pH} = 4.74 + \log_{10}\left(\frac{0.2}{0.1}\right) = 4.74 + \log_{10}(2) = 4.74 + 0.30 = 5.04$

Worked Example 13

What is the pH of the buffer after adding $0.01 \mathrm{ mol}$ of $\mathrm{HCl}$ to $1 \mathrm{ dm}^3$ of the buffer in Worked Example 12?

$\mathrm{HCl}$ reacts with $\mathrm{CH}_3\mathrm{COO}^-$:

New $[\mathrm{CH}_3\mathrm{COO}^-] = 0.2 - 0.01 = 0.19 \mathrm{ mol/dm}^3$

New $[\mathrm{CH}_3\mathrm{COOH}] = 0.1 + 0.01 = 0.11 \mathrm{ mol/dm}^3$

$\mathrm{pH} = 4.74 + \log_{10}\left(\frac{0.19}{0.11}\right) = 4.74 + \log_{10}(1.727) = 4.74 + 0.237 = 4.98$

The pH changed from 5.04 to 4.98, a change of only 0.06. Without the buffer, adding $0.01 \mathrm{ mol}$ of $\mathrm{HCl}$ to $1 \mathrm{ dm}^3$ of water would give pH = 2.

:::info Info $\mathrm{pH} = \mathrm{p}K_a$. The effective buffering range is approximately $\mathrm{p}K_a \pm 1$.

Indicators and pH Curves

Strong Acid - Strong Base Titration

• Equivalence point at pH = 7
• Sharp change in pH around the equivalence point
• Suitable indicators: phenolphthalein, methyl orange, universal indicator

Weak Acid - Strong Base Titration

• Equivalence point at pH \gt 7 (salt hydrolysis produces $\mathrm{OH}^-$)
• Less sharp change in pH
• Suitable indicator: phenolphthalein (pH range 8.3-10.0)

Strong Acid - Weak Base Titration

• Equivalence point at pH \lt 7 (salt hydrolysis produces $\mathrm{H}^+$)
• Suitable indicator: methyl orange (pH range 3.1-4.4)

Worked Example 14

$25.0 \mathrm{ cm}^3$ of $0.100 \mathrm{ mol/dm}^3$ ethanoic acid ($\mathrm{CH}_3\mathrm{COOH}$ $K_a = 1.8 \times 10^{-5}$) is titrated with $0.100 \mathrm{ mol/dm}^3$ $\mathrm{NaOH}$. Find the pH At the equivalence point.

At equivalence point: moles of $\mathrm{NaOH}$ = moles of $\mathrm{CH}_3\mathrm{COOH}$

Moles of $\mathrm{CH}_3\mathrm{COOH} = 0.100 \times 0.0250 = 0.00250 \mathrm{ mol}$

Volume of $\mathrm{NaOH}$ needed = $0.0250 \mathrm{ dm}^3 = 25.0 \mathrm{ cm}^3$

Total volume = $50.0 \mathrm{ cm}^3 = 0.0500 \mathrm{ dm}^3$

The salt $\mathrm{CH}_3\mathrm{COONa}$ is formed. Concentration of $\mathrm{CH}_3\mathrm{COO}^-$:

$[\mathrm{CH}_3\mathrm{COO}^-] = \frac{0.00250}{0.0500} = 0.0500 \mathrm{ mol/dm}^3$

The $\mathrm{CH}_3\mathrm{COO}^-$ ion hydrolyses:

$\mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{CH}_3\mathrm{COOH} + \mathrm{OH}^-$

$K_b = \frac{K_w}{K_a} = \frac{10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$

$K_b = \frac{[\mathrm{CH}_3\mathrm{COOH}][\mathrm{OH}^-]}{[\mathrm{CH}_3\mathrm{COO}^-]} = \frac{x^2}{0.0500} = 5.56 \times 10^{-10}$

$x^2 = 2.78 \times 10^{-11}$

$x = 5.27 \times 10^{-6} \mathrm{ mol/dm}^3$

$\mathrm{pOH} = -\log_{10}(5.27 \times 10^{-6}) = 5.28$

$\mathrm{pH} = 14 - 5.28 = 8.72$

Corrosion and Its Prevention

Rusting of Iron

Rusting is an electrochemical process that requires both water and oxygen.

At the anode (oxidation):

$\mathrm{Fe} \to \mathrm{Fe}^{2+} + 2e^-$

At the cathode (reduction):

$\mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O} + 4e^- \to 4\mathrm{OH}^-$

The $\mathrm{Fe}^{2+}$ reacts with $\mathrm{OH}^-$ and oxygen to form hydrated iron(III) oxide (rust):

$4\mathrm{Fe}^{2+} + 4\mathrm{OH}^- + \mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O} \to 4\mathrm{Fe}(\mathrm{OH})_3$

Methods of Prevention

Additional Worked Examples

Worked Example: Buffer pH Calculation

Calculate the pH of a buffer containing $0.20 \mathrm{ mol/dm^3}$ $\mathrm{CH_3COOH}$ ($K_a = 1.8 \times 10^{-5}$) and $0.10 \mathrm{ mol/dm^3}$ $\mathrm{CH_3COONa}$.

$\mathrm{p}K_a = -\log_{10}(1.8 \times 10^{-5}) = 4.74$

Using the Henderson-Hasselbalch equation:

$\mathrm{pH} = \mathrm{p}K_a + \log_{10}\left(\frac{[\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]}\right) = 4.74 + \log_{10}\left(\frac{0.10}{0.20}\right) = 4.74 + \log_{10}(0.5) = 4.74 - 0.30 = 4.44$

Worked Example: Electrolysis Product Prediction

Predict the products of electrolysis of dilute $\mathrm{Na_2SO_4}$ solution using inert carbon electrodes.

Ions present: \mathrm{Na^+}$$\mathrm{SO_4^{2-}}$$\mathrm{H^+}$$\mathrm{OH^-}

Cathode: $\mathrm{Na^+}$ is above $\mathrm{H^+}$ in the discharge series, so water is reduced:

$2\mathrm{H_2O} + 2e^- \to \mathrm{H_2} + 2\mathrm{OH^-}$

Anode: $\mathrm{SO_4^{2-}}$ is not a halide, so water is oxidised:

$4\mathrm{OH^-} \to \mathrm{O_2} + 2\mathrm{H_2O} + 4e^-$

Products: hydrogen gas (cathode) and oxygen gas (anode). The solution becomes increasingly alkaline due to \mathrm{OH^- accumulation.

Worked Example: Choosing a Salt Preparation Method

Describe how to prepare pure, dry crystals of zinc sulphate.

$\mathrm{ZnSO_4}$ is a soluble salt. Zinc is a reactive metal (above hydrogen), so the acid + metal method can be used:

1. Add excess zinc granules to dilute $\mathrm{H_2SO_4}$ in a beaker.
2. Effervescence occurs as $\mathrm{H_2}$ is produced: $\mathrm{Zn} + \mathrm{H_2SO_4} \to \mathrm{ZnSO_4} + \mathrm{H_2}$
3. Continue until no more gas is produced (excess zinc ensures all acid is consumed).
4. Filter to remove excess zinc.
5. Heat the filtrate to concentrate by evaporation until crystals begin to form.
6. Allow to cool for crystallisation, then filter and dry.

Additional Practice Questions

Question 6: Calculate the pH of a buffer made by mixing $100 \mathrm{ cm}^3$ of $0.20 \mathrm{ mol/dm}^3$ $\mathrm{NH}_3$ ($K_b = 1.8 \times 10^{-5}$) with $100 \mathrm{ cm}^3$ of $0.10 \mathrm{ mol/dm}^3$ $\mathrm{HCl}$.

$\mathrm{NH}_3$ reacts with $\mathrm{HCl}$: $\mathrm{NH}_3 + \mathrm{HCl} \to \mathrm{NH}_4^+ + \mathrm{Cl}^-$

Moles of $\mathrm{NH}_3 = 0.20 \times 0.100 = 0.0200 \mathrm{ mol}$

Moles of $\mathrm{HCl} = 0.10 \times 0.100 = 0.0100 \mathrm{ mol}$

Remaining $\mathrm{NH}_3 = 0.0200 - 0.0100 = 0.0100 \mathrm{ mol}$

$\mathrm{NH}_4^+$ formed $= 0.0100 \mathrm{ mol}$

Total volume $= 0.200 \mathrm{ dm}^3$

$[\mathrm{NH}_3] = 0.0500 \mathrm{ mol/dm}^3$, $[\mathrm{NH}_4^+] = 0.0500 \mathrm{ mol/dm}^3$

$\mathrm{p}K_a = 14 - \mathrm{p}K_b = 14 - (-\log_{10}(1.8 \times 10^{-5})) = 14 - 4.74 = 9.26$

$\mathrm{pH} = 9.26 + \log_{10}\left(\frac{0.0500}{0.0500}\right) = 9.26 + 0 = 9.26$

Question 7: Explain why steel wool rusts faster when in contact with less active metals like Copper.

Copper is less reactive than iron. When in contact, iron acts as the anode and copper as the Cathode. The iron oxidises more rapidly, accelerating the rusting process. This is an example of electrochemical corrosion where the less reactive metal provides a surface for the reduction Reaction.

Question 8: A current of $4.0 \mathrm{ A}$ is passed through $\mathrm{CuSO}_4$ solution using Copper electrodes. The mass of the anode decreased by $2.38 \mathrm{ g}$ after a certain time. Calculate the time for which the current was passed.

$m = \frac{Q \times M}{nF}$

$2.38 = \frac{Q \times 63.5}{2 \times 96500}$

$Q = \frac{2.38 \times 2 \times 96500}{63.5} = \frac{459340}{63.5} = 7234 \mathrm{ C}$

$t = \frac{Q}{I} = \frac{7234}{4.0} = 1808.5 \mathrm{ s} \approx 30.1 \mathrm{ minutes}$

Problem Set

Problem 1: Calculate the pH of $0.0025 \mathrm{ mol/dm^3}$ $\mathrm{H_2SO_4}$.

If you get this wrong, revise: The pH Scale

Details

$\mathrm{H_2SO_4}$ is a strong diprotic acid: $[\mathrm{H^+}] = 2 \times 0.0025 = 0.0050 \mathrm{ mol/dm^3}$

$\mathrm{pH} = -\log_{10}(0.0050) = 2.30$

Problem 2: Calculate the pH of $0.050 \mathrm{ mol/dm^3}$ $\mathrm{CH_3COOH}$ ($K_a = 1.8 \times 10^{-5}$).

If you get this wrong, revise: Strong and Weak Acids

$K_a = \frac{x^2}{0.050} = 1.8 \times 10^{-5}$

$x = \sqrt{9.0 \times 10^{-7}} = 9.49 \times 10^{-4} \mathrm{ mol/dm^3}$

$\mathrm{pH} = -\log_{10}(9.49 \times 10^{-4}) = 3.02$

Problem 3: A buffer contains $0.15 \mathrm{ mol/dm^3}$ $\mathrm{NH_3}$ and $0.15 \mathrm{ mol/dm^3}$ $\mathrm{NH_4Cl}$ ($K_b = 1.8 \times 10^{-5}$). Calculate the pH.

If you get this wrong, revise: Buffers

$\mathrm{p}K_b = -\log_{10}(1.8 \times 10^{-5}) = 4.74$

$\mathrm{p}K_a = 14 - 4.74 = 9.26$

$\mathrm{pH} = \mathrm{p}K_a + \log_{10}\left(\frac{[\mathrm{NH_3}]}{[\mathrm{NH_4^+}]}\right) = 9.26 + \log_{10}\left(\frac{0.15}{0.15}\right) = 9.26 + 0 = 9.26$

Problem 4: $25.0 \mathrm{ cm^3}$ of $0.100 \mathrm{ mol/dm^3}$ $\mathrm{HCl}$ is titrated with $0.0800 \mathrm{ mol/dm^3}$ $\mathrm{NaOH}$. Calculate the volume of $\mathrm{NaOH}$ needed to reach the equivalence point.

If you get this wrong, revise: Acid-Base Titrations

$\mathrm{HCl} + \mathrm{NaOH} \to \mathrm{NaCl} + \mathrm{H_2O}$

$n(\mathrm{HCl}) = 0.100 \times 0.0250 = 0.00250 \mathrm{ mol}$

$V(\mathrm{NaOH}) = \frac{n}{c} = \frac{0.00250}{0.0800} = 0.03125 \mathrm{ dm^3} = 31.3 \mathrm{ cm^3}$

Problem 5: Describe how to prepare pure, dry crystals of lead(II) nitrate.

If you get this wrong, revise: Salt Preparation

$\mathrm{Pb(NO_3)_2}$ is soluble, and lead is below hydrogen in the reactivity series (it does not react with dilute acids to produce $\mathrm{H_2}$). The best method is acid + insoluble base:

1. Add excess $\mathrm{PbO}$ or $\mathrm{PbCO_3}$ to dilute $\mathrm{HNO_3}$
2. $\mathrm{PbO} + 2\mathrm{HNO_3} \to \mathrm{Pb(NO_3)_2} + \mathrm{H_2O}$
3. Filter to remove excess base
4. Evaporate filtrate to crystallisation
5. Filter, wash, and dry

Problem 6: Assign oxidation numbers to all elements in $\mathrm{H_2O_2}$ (hydrogen peroxide).

If you get this wrong, revise: Oxidation Numbers

$\mathrm{H_2O_2}$ is a peroxide. In peroxides, oxygen has oxidation number $-1$.

$2(+1) + 2(-1) = 0$

$\mathrm{H} = +1$, $\mathrm{O} = -1$.

This is an exception to the usual rule that oxygen is $-2$.

Problem 7: Balance the reaction of $\mathrm{MnO_4^-}$ with $\mathrm{H_2S}$ in acidic solution to give $\mathrm{Mn^{2+}}$ and $\mathrm{S}$.

If you get this wrong, revise: Balancing Redox Equations

Reduction: $\mathrm{MnO_4^-} + 8\mathrm{H^+} + 5e^- \to \mathrm{Mn^{2+}} + 4\mathrm{H_2O}$

Oxidation: $\mathrm{H_2S} \to \mathrm{S} + 2\mathrm{H^+} + 2e^-$

Multiply reduction by 2 and oxidation by 5:

$2\mathrm{MnO_4^-} + 16\mathrm{H^+} + 10e^- \to 2\mathrm{Mn^{2+}} + 8\mathrm{H_2O}$

$5\mathrm{H_2S} \to 5\mathrm{S} + 10\mathrm{H^+} + 10e^-$

Overall: $2\mathrm{MnO_4^-} + 5\mathrm{H_2S} + 6\mathrm{H^+} \to 2\mathrm{Mn^{2+}} + 5\mathrm{S} + 8\mathrm{H_2O}$

Problem 8: Predict the products at each electrode when concentrated aqueous $\mathrm{NaCl}$ is electrolysed using carbon electrodes. Write half-equations and the overall equation.

If you get this wrong, revise: Electrolysis of Aqueous Solutions

Cathode: $\mathrm{Na^+}$ is above $\mathrm{H^+}$So $\mathrm{H_2}$ is produced:

$2\mathrm{H_2O} + 2e^- \to \mathrm{H_2} + 2\mathrm{OH^-}$

Anode: Concentrated $\mathrm{Cl^-}$ is discharged:

$2\mathrm{Cl^-} \to \mathrm{Cl_2} + 2e^-$

Overall: $2\mathrm{H_2O} + 2\mathrm{NaCl} \to \mathrm{H_2} + \mathrm{Cl_2} + 2\mathrm{NaOH}$

Problem 9: What mass of silver is deposited when a current of $0.60 \mathrm{ A}$ is passed through $\mathrm{AgNO_3}$ solution for 25 minutes?

If you get this wrong, revise: Faraday’s Laws of Electrolysis

$Q = 0.60 \times 25 \times 60 = 900 \mathrm{ C}$

$\mathrm{Ag^+} + e^- \to \mathrm{Ag} \quad (n = 1)$

$m = \frac{Q \times M}{nF} = \frac{900 \times 108}{1 \times 96500} = \frac{97200}{96500} = 1.01 \mathrm{ g}$

Problem 10: A cell is made from $\mathrm{Mg}^{2+}/\mathrm{Mg}$ ($E^\circ = -2.37 \mathrm{ V}$) and $\mathrm{Ni}^{2+}/\mathrm{Ni}$ ($E^\circ = -0.25 \mathrm{ V}$). Calculate $E^\circ_{\mathrm{cell}}$ and write the overall equation. Is the reaction spontaneous?

If you get this wrong, revise: Electrochemical Cells

Nickel has the more positive $E^\circ$ (cathode, reduction).

$E^\circ_{\mathrm{cell}} = -0.25 - (-2.37) = 2.12 \mathrm{ V}$