Chemistry - Acids, Bases, and Electrochemistry

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

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 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}), $\mathrm{Ba(OH)}_2$

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

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

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 $25^\circ\mathrm{C}$.

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:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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)

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

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

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 $0.10 \mathrm{ cm}^3$)

$25.0 \mathrm{ cm}^3$ 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 $\mathrm{NaOH} = 0.00200 \mathrm{ mol}$

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

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

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

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

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

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.

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

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: 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

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.

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

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)

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

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

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 $= 96500 \mathrm{ C/mol}$

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

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

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

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

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

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. ($1 \mathrm{ mol} = 24.0 \mathrm{ dm}^3$):

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

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

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.

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 $E^\circ = 0.00 \mathrm{ V}$.

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

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.

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 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).

The proton shuffle: Acids donate protons (H⁺), bases accept them — it’s like a game of hot potato where protons are passed between molecules. pH measures how many free protons are floating around.

Why it matters: From stomach acid to blood buffers, acid-base chemistry keeps biological systems alive. Understanding pH helps design medicines, treat water, and control industrial processes.

The key insight: Buffers resist pH change by absorbing or releasing protons — they’re the body’s way of maintaining equilibrium despite constant disturbances.

flowchart TD
    A[1_Acids Bases And Electrochemistry] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

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

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

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)

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 $\mathrm{OH}^-$.

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

Where:

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

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.

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

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

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.

• Atomic Structure: Atomic structure determines bonding
• Equilibrium: Equilibrium is a core topic
• Organic Chemistry: Organic chemistry covers carbon compounds