Class 11 Chemistry – Chapter 9: Acids, Bases & Salts (FBISE)
This section provides complete, exam-oriented notes for Class 11 Chemistry Chapter 9 – Acids, Bases & Salts strictly according to the Federal Board (FBISE) syllabus. Concepts are explained clearly to build strong fundamentals.
Key topics include Arrhenius, Brønsted–Lowry and Lewis concepts of acids and bases, pH scale, strength of acids and bases, buffers, common salts, hydrolysis of salts, and everyday applications. Important reactions, tables, and exam-focused points are highlighted.
For better understanding and exam preparation, students can access video lectures, MCQs practice, numericals, test series, and live revision sessions on our YouTube channel and stay connected through our WhatsApp channel.
1. Introduction to Acids and Bases
Acids and bases are fundamental substances in chemistry, often identified by their physical properties and chemical behavior.
- Acids: Identified by a sour taste (e.g., acetic acid in vinegar, citric acid in lemons). They are classified into mineral acids and organic acids (which are generally weaker).
- Bases (Alkalis): Characterized by a bitter taste and slippery feel. Soluble bases are specifically called alkalis and release hydroxide ions ($OH^-$) in water.
- Salts: Ionic compounds formed through the neutralization reaction between an acid and a base.
Common Organic Acids and Their Sources
| Organic Acid | Occurrence |
|---|---|
| Lactic acid | Sour Milk |
| Citric acid | Citrus fruits (lemons, oranges) |
| Formic acid | Insect bites |
| Tartaric acid | Grape juice |
| Maleic acid | Apples and pears |
2. Chemical Reactions and Ionization
The behavior of acids and bases often depends on the presence of water ($H_2O$) to facilitate the formation of ions.
Acid Ionization in Water:
Pure $HCl$ is covalent, but in water, it forms hydronium ions:
$$HCl + H_2O \longrightarrow H_3O^+ + Cl^-$$
Base Dissociation:
A metal oxide or hydroxide reacting with water to form an alkali:
$$CaO + H_2O \longrightarrow Ca(OH)_2$$
Neutralization Reaction:
$$HCl + NaOH \longrightarrow NaCl + H_2O$$
(Acid + Base $\longrightarrow$ Salt + Water)
3. Brønsted-Lowry Conjugate Acid-Base Pairs
According to this theory, acids and bases work in pairs based on proton ($H^+$) transfer.
- Conjugate Acid: The species formed when a base accepts a proton.
- Conjugate Base: The species remaining after an acid donates a proton.
The Inverse Relationship: A strong acid produces a relatively weak conjugate base. Conversely, a strong base produces a relatively weak conjugate acid.
Relevant Questions and Answers
Q1: Why is water essential for $HCl$ to behave as an acid?
A: $HCl$ is covalent in nature and does not form $H^+$ ions on its own. In the presence of $H_2O$, it reacts to form hydronium ions ($H_3O^+$), which gives it acidic properties.
Q2: Define an Alkali.
A: An alkali is a soluble base that dissociates in water to form hydroxide ions ($OH^-$).
Q3: In the reaction $$NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$$, identify the conjugate acid-base pairs.
A:
- $NH_3$ (Base) and $NH_4^+$ (Conjugate Acid)
- $H_2O$ (Acid) and $OH^-$ (Conjugate Base)
Q4: What is the difference between a mineral acid and an organic acid?
A: Mineral acids are typically inorganic and much stronger, whereas organic acids are derived from natural sources (like fruits and vegetables) and are much weaker.
Q5: How do conjugate acid-base pairs differ structurally?
A: They differ by exactly one proton ($H^+$).
9.2 Lewis Concept of Acids and Bases
The Lewis concept, proposed by G.N. Lewis in 1923, is a more general theory than the Arrhenius or Brønsted-Lowry theories. It focuses on the exchange of electron pairs rather than proton transfer.
- Lewis Acid: A substance that can accept a pair of electrons to form a coordinate covalent bond.
- Lewis Base: A substance that can donate a pair of electrons to form a coordinate covalent bond.
In a Lewis acid-base reaction, a coordinate covalent bond (or dative bond) is formed between the donor (base) and the acceptor (acid).
Key Characteristics
| Feature | Lewis Acid | Lewis Base |
|---|---|---|
| Electron Status | Deficient in electrons (Incomplete octet) | Possesses at least one unshared (lone) pair of electrons |
| Role | Electron pair acceptor | Electron pair donor |
| Example | $BF_{3}$, $AlCl_{3}$, $H^{+}$ | $NH_{3}$, $OH^{-}$, $Cl^{-}$ |
9.3 Strength of Acids and Bases
The strength of Brønsted acids and bases is determined by their ability to donate or accept protons ($H^{+}$):
- Acid Strength: A "strong" acid donates protons to a higher degree than a "weak" acid. For example, $HCl$ is a stronger acid than acetic acid.
- Base Strength: A "strong" base accepts protons more readily than a "weak" base. Ammonia ($NH_{3}$) is a stronger base than water ($H_{2}O$).
Concept Assessment & Practice
Solution to Concept Assessment Exercise 9.1
Identify the Lewis acid and Lewis base in the following reactions:
1. $Cl^{-} + AlCl_{3} \longrightarrow [AlCl_{4}]^{-}$
- Lewis Base: $Cl^{-}$ (It has lone pairs and donates an electron pair).
- Lewis Acid: $AlCl_{3}$ (The Aluminum atom has an incomplete octet and accepts the pair).
2. $H^{+} + OH^{-} \longrightarrow H_{2}O$
- Lewis Base: $OH^{-}$ (The oxygen atom donates a lone pair).
- Lewis Acid: $H^{+}$ (The proton accepts the electron pair to fill its valence shell).
Relevant Questions & Answers
Q1: Why is $BF_{3}$ considered a Lewis acid?
A: Boron in $BF_{3}$ has only six electrons in its valence shell (an incomplete octet). Therefore, it needs an electron pair to complete its octet, making it an electron pair acceptor.
Q2: What is the main difference between a Brønsted base and a Lewis base?
A: A Brønsted base is defined by its ability to accept a proton ($H^{+}$), whereas a Lewis base is defined by its ability to donate an electron pair. In many cases, like $NH_{3}$, a substance can be both.
Q3: What type of bond is formed during a Lewis acid-base reaction?
A: A coordinate covalent bond is formed, where both electrons in the shared pair originate from the Lewis base.
9.4 Ionisation Constant of Water and Calculation of pH
Water is an amphoteric compound, meaning it can act as both a Brønsted acid (proton donor) and a Brønsted base (proton acceptor). This unique property leads to auto-ionization.
Key Concepts & Notes
- Auto-ionization Equation: Water molecules react with each other to form hydronium and hydroxide ions:
$$H_2O + H_2O \rightleftharpoons H_3O^+ + OH^-$$
- The Ionic Product of Water ($$K_w$$): Since the concentration of pure water is constant, the equilibrium constant is simplified to:
$$K_w = [H^+][OH^-]$$
- Standard Value: At 25°C, the value of $$K_w$$is exactly$$1.0 \times 10^{-14}$$.
- Temperature Dependence: $$K_w$$is an equilibrium constant and changes with temperature. For example, at 40°C,$$K_w$$increases to$$3.8 \times 10^{-14}$$.
- pH and pOH Definition:
- $$pH = -\log[H^+]$$
- $$pOH = -\log[OH^-]$$
- The relationship between them is: $$pH + pOH = 14$$ (at 25°C).
Solution Characterization (at 25°C)
| Solution Type | Ion Concentration Relationship | pH Range |
|---|---|---|
| Neutral | $$[H^+] = [OH^-] = 1.0 \times 10^{-7} M$$ | $$pH = 7.00$$ |
| Acidic | $$[H^+] > [OH^-]$$or$$[H^+] > 1.0 \times 10^{-7} M$$ | $$pH < 7.00$$ |
| Basic | $$[OH^-] > [H^+]$$or$$[H^+] < 1.0 \times 10^{-7} M$$ | $$pH > 7.00$$ |
Questions & Answers
Q1: Why is the concentration of water $$[H_2O]$$omitted from the final$$K_w$$ expression?
A: Water is a weak electrolyte and acts as the solvent. Because it is in such a large excess, its concentration remains effectively constant during the ionization process. Therefore, it is incorporated into the constant $$K$$to create the new constant$$K_w$$.
Q2: If the temperature of pure water is increased to 40°C, does it become acidic?
A: No. Although the concentration of $$[H^+]$$increases to$$1.9 \times 10^{-7} M$$(making the pH lower than 7), the concentration of$$[OH^-]$$increases by the exact same amount. Since$$[H^+] = [OH^-]$$, the water remains neutral.
Q3: Derive the relationship between pH and pOH from the $$K_w$$ expression.
A: Starting with $$K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$$.Taking the negative log of both sides:$$-\log(1.0 \times 10^{-14}) = -\log([H^+][OH^-])$$$$14 = (-\log[H^+]) + (-\log[OH^-])$$$$14 = pH + pOH$$
Q4: What is the pH of a solution where $$[OH^-] = 1.0 \times 10^{-4} M$$ at 25°C?
A: First, find pOH: $$pOH = -\log(1.0 \times 10^{-4}) = 4$$.Using the relation $$pH + pOH = 14$$:$$pH = 14 - 4 = 10$$.The solution is basic.
Notes: pH and pOH Calculations
1. Core Concepts and Formulas
- Ion Product Constant of Water ($K_w$): At 25°C, $K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$. This relationship allows for the calculation of $[H^+]$ if $[OH^-]$ is known, and vice versa.
- pH Definition: The negative logarithm of the hydrogen ion concentration: $pH = -\log[H^+]$.
- pOH Definition: The negative logarithm of the hydroxide ion concentration: $pOH = -\log[OH^-]$.
- The Relationship: For any aqueous solution, $pH + pOH = 14$.
2. Ionization of Acids and Bases
- Strong Acids (e.g., $HCl$): Ionize completely in water. For a $0.001 M$ solution of $HCl$, the $[H^+]$ is also $0.001 M$ because the ratio is 1:1.
- Strong Bases (e.g., $NaOH$): Dissociate completely. A $0.062 M$ solution of $NaOH$ yields a $[OH^-]$ of $0.062 M$.
- Polyprotic Acids (e.g., $H_2SO_4$): These acids can donate more than one proton. In the case of sulphuric acid, one mole of acid provides two moles of $H^+$ ions upon complete ionization.
Practice Questions and Solutions
Question 1: A household ammonia solution has a $[OH^-]$ of $0.005 M$. What is the concentration of $[H^+]$?
Answer:Using the formula $K_w = [H^+][OH^-]$:$1.0 \times 10^{-14} = [H^+] \times 0.005$$[H^+] = \frac{1.0 \times 10^{-14}}{0.005} = 2.0 \times 10^{-12} M$
Question 2: Calculate the pH of a $0.062 M$ $NaOH$ solution.
Answer:1. Find pOH: $pOH = -\log[0.062] \approx 1.21$2. Use the relation $pH + pOH = 14$:$pH = 14 - 1.21 = 12.79$
Question 3 (Concept Assessment 9.2): What is the pH of a solution containing $1.95g$ pure $H_2SO_4$ per $dm^3$ of solution?
Answer:Step 1: Calculate Molarity of $H_2SO_4$Molar mass of $H_2SO_4 = 2(1.0) + 32.0 + 4(16.0) = 98 g/mol$$Moles = \frac{1.95g}{98 g/mol} \approx 0.0199 mol$Since the volume is $1 dm^3$, $Molarity = 0.0199 M$.Step 2: Calculate $[H^+]$$H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$$[H^+] = 2 \times 0.0199 M = 0.0398 M$Step 3: Calculate pH$pH = -\log(0.0398) \approx 1.4$
Notes: Acid-Base Titrations
Definition: Titration is a volumetric method used to determine the concentration of an unknown solution (the analyte) by reacting it completely with a solution of known concentration (the standard solution).
1. Experimental Procedure
- Apparatus: Burettes and volumetric pipettes are essential.
- Setup: Generally, the acid is placed in the burette, while a fixed volume of base is placed in a conical flask.
- Indicator: A few drops of an acid-base indicator are added to the flask to signal the end point through a color change.
- Neutralization: For a strong acid-strong base titration using phenolphthalein, the base solution starts pink and becomes colorless at the end point.
2. Mathematical Calculation
The molarity of the unknown solution is calculated using the titration equation:
$$\frac{M_1 V_1}{n_1} = \frac{M_2 V_2}{n_2}$$
- $M_1$: Molarity of the base
- $V_1$: Volume of the base in the flask
- $n_1$: Number of moles of base from the balanced equation
- $M_2$: Molarity of the acid
- $V_2$: Volume of acid used from the burette
- $n_2$: Number of moles of acid from the balanced equation
Strength of solution: To find the strength in $g/dm^3$, use the formula: $$\text{Strength} = \text{Molarity} \times \text{Molar mass}$$
3. Selection of Indicators
An indicator is chosen based on its pH range and the nature of the reactants:
| Reactants | pH Range at End Point | Recommended Indicator |
|---|---|---|
| Strong Acid + Strong Base | $3$ to $11$ | Phenolphthalein (or Methyl Orange) |
| Weak Acid + Strong Base | $6.5$ to $10$ | Phenolphthalein |
| Strong Acid + Weak Base | $3.5$ to $7.5$ | Methyl Orange (or Bromocresol Green) |
| Weak Acid + Weak Base | No sharp change | None works satisfactorily |
Questions and Answers
Q1: What is the "end point" in a titration?
A: The end point is the physical point in a titration where the indicator changes color, signaling that the reaction between the two solutions is complete.
Q2: Why is phenolphthalein a good choice for a titration between a weak acid (like oxalic acid) and a strong base (sodium hydroxide)?
A: In a weak acid-strong base titration, the pH changes within the range of $6.5$ to $10$. Since phenolphthalein has an application range of pH $8.3$ to $10$, it falls within this limit and can accurately indicate the end point.
Q3: Calculate the molarity of $HCl$ if $30 cm^3$ of it neutralizes $25 cm^3$ of $0.12M$ $NaOH$.
A: Given the reaction: $HCl + NaOH \rightarrow NaCl + H_2O$, where $n_1 (base) = 1$ and $n_2 (acid) = 1$.Using $$\frac{M_1 V_1}{n_1} = \frac{M_2 V_2}{n_2}$$:$$\frac{0.12 \times 25}{1} = \frac{M_2 \times 30}{1}$$$$M_2 = \frac{0.12 \times 25}{30} = 0.1M$$The molarity of the $HCl$ solution is $0.1M$.
Q4: Why does sherbet fizz on the tongue?
A: Sherbet contains citric acid and baking soda. When eaten, the acid dissolves in saliva and reacts with the baking soda (a base) in a neutralization reaction. This reaction produces carbon dioxide gas, which creates the fizzing sensation.
9.6 Strong and Weak Acids: Study Notes
1. Acid Dissociation Constant ($K_a$)
The strength of an acid is determined by its extent of ionization in water. For a general acid $HX$, the equilibrium is:
$HX_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + X^-_{(aq)}$
The equilibrium constant $K_a$ (acid dissociation constant) is defined as:
$K_a = \frac{[H_3O^+][X^-]}{[HX]}$
- Greater $K_a$ = Stronger acid (more ionization).
- Smaller $K_a$ = Weaker acid (less ionization).
- $K_a$ is temperature-dependent; values are typically cited at $25^\circ\text{C}$.
2. The $pK_a$ Scale
To handle the wide range of $K_a$ values, the $pK_a$ scale is used:
$pK_a = -\log K_a$
Relationship Summary:
| Acid Strength | $K_a$ Value | $pK_a$ Value |
|---|---|---|
| Stronger | Higher (e.g., $1.0 \times 10^{10}$) | Lower/Negative (e.g., $-10.0$) |
| Weaker | Lower (e.g., $1.8 \times 10^{-16}$) | Higher/Positive (e.g., $+15.7$) |
3. Simplified Calculation for Weak Acids
When calculating $[H^+]$ for a weak acid where ionization ($x$) is very small compared to the initial concentration (e.g., $1.0\text{M}$):
- Assume $1.0 - x \approx 1.0$.
- Formula: $K_a = \frac{x^2}{Initial Concentration}$.
- Example for $1.0\text{M HF}$: $x = \sqrt{K_a \times 1.0} = \sqrt{7.2 \times 10^{-4}} = 0.0268\text{M}$.
Relevant Questions & Answers
A: The strongest is Perchloric acid ($HClO_4$) with a $pK_a$ of $-10.0$. The weakest is Water ($H_2O$) with a $pK_a$ of $+15.7$.
A: $HCl$ has a much higher $K_a$ ($1.0 \times 10^6$) compared to $HF$ ($7.2 \times 10^{-4}$). This means $HCl$ dissociates almost completely in water, while $HF$ only dissociates partially.
A: Since the value is positive and relatively small, it indicates that Acetic acid is a weak acid. It exists primarily in its molecular form ($CH_3COOH$) in solution rather than as ions.
A: For weak acids, the value of $x$ (the amount dissociated) is so small that subtracting it from the initial concentration does not significantly change the value within the limits of significant figures.
Notes: Buffer Solutions and Bases
1. Strong and Weak Bases
- Base Strength: The ability of a base to accept a proton ($H^{+}$) from a solvent.
- Strong Bases: Hydroxides of alkali metals (e.g., $NaOH$, $KOH$) that ionize completely in aqueous solutions.
- Base Ionization Constant ($K_{b}$): Represents the equilibrium constant for a base reacting with water:
$$B + H_{2}O_{(aq)} \rightleftharpoons BH^{+}_{(aq)} + OH^{-}_{(aq)}$$
$$K_{b} = \frac{[BH^{+}][OH^{-}]}{[B]}$$ - Relationship: A large $K_{b}$ indicates a strong base (high ionization), while a small $K_{b}$ indicates a weak base.
- $pK_{b}$: Defined as the negative logarithm of $K_{b}$: $$pK_{b} = -\log K_{b}$$
2. Buffer Solutions
- Definition: A solution that resists significant changes in $pH$ when small amounts of acid or base are added.
- Types of Buffers:
- Acidic Buffers: Made by mixing a weak acid and its salt with a strong base (e.g., $CH_{3}COOH + CH_{3}COONa$). $pH < 7$.
- Basic Buffers: Made by mixing a weak base and its salt with a strong acid (e.g., $NH_{4}OH + NH_{4}Cl$). $pH > 7$.
3. Henderson-Hasselbalch Equation
Used to calculate the $pH$ or $pOH$ of a buffer solution:
- For Acidic Buffers: $$pH = pK_{a} + \log \frac{[salt]}{[acid]}$$
- For Basic Buffers: $$pOH = pK_{b} + \log \frac{[salt]}{[base]}$$
4. Biological $pH$ Ranges
| Biological System | $pH$ Range |
|---|---|
| Human blood | 7.35 to 7.45 |
| Tears | 7.40 |
| Stomach | 1.65 - 1.75 |
| Milk | 6.7 - 6.8 |
Concept Assessment Questions and Answers
Question 1: Calculate the $pH$ of a buffer solution in which $0.11 \text{ M } CH_{3}COONa$ and $0.09 \text{ M } CH_{3}COOH$ solutions are present. ($K_{a}$ for $CH_{3}COOH$ is $1.8 \times 10^{-5}$)
Answer:
- Find $pK_{a}$:
$$pK_{a} = -\log K_{a}$$
$$pK_{a} = -\log(1.8 \times 10^{-5}) \approx 4.74$$ - Apply Henderson-Hasselbalch Equation:
$$pH = pK_{a} + \log \frac{[salt]}{[acid]}$$
$$pH = 4.74 + \log \frac{0.11}{0.09}$$
$$pH = 4.74 + \log(1.222)$$
$$pH = 4.74 + 0.087 = 4.827 \approx 4.83$$
Question 2: What is the $pH$ of a buffer if $[CH_{3}COOH] = 0.1 \text{ M}$ and $[CH_{3}COONa] = 1.0 \text{ M}$ with a $pK_{a}$ of 4.76?
Answer:
$$pH = 4.76 + \log \frac{1.0}{0.1}$$
$$pH = 4.76 + \log(10)$$
$$pH = 4.76 + 1.00 = 5.76$$
Question 3: Why is the $OH^{-}$ ion considered a Brönsted base?
Answer: The $OH^{-}$ ion is considered a Brönsted base because it has the ability to accept a proton ($H^{+}$) to form water ($H_{2}O$):
$$OH^{-} + H^{+} \rightarrow H_{2}O$$
Notes: Salt Hydrolysis
1. Definition of Hydrolysis
Hydrolysis is the chemical reaction of cations or anions (or both) of a salt with water. It is distinct from hydration because hydrolysis involves the breaking of $H-OH$ bonds, whereas hydration is simply the addition of water molecules to a substance without bond cleavage.
2. Mechanism based on Bronsted-Lowry Theory
When a salt $MX$ dissolves, it dissociates into $M^+$ (cation) and $X^-$ (anion). These ions can react with water as follows:
- Cation Reaction: $M^+_{(aq)} + H-OH_{(l)} \rightleftharpoons MOH_{(aq)} + H^+_{(aq)}$ (Increases acidity)
- Anion Reaction: $X^-_{(aq)} + H-OH_{(l)} \rightleftharpoons HX_{(aq)} + OH^-_{(aq)}$ (Increases basicity)
3. Types of Salt Hydrolysis
| Acid Parent | Base Parent | Hydrolysis Behavior | Solution pH | Examples |
|---|---|---|---|---|
| Strong | Strong | No hydrolysis occurs. | $7.0$ (Neutral) | $NaCl$, $KNO_3$, $Na_2SO_4$ |
| Strong | Weak | Cations react with water. | $< 7.0$ (Acidic) | $NH_4Cl$, $CuSO_4$, $NH_4NO_3$ |
| Weak | Strong | Anions react with water. | $> 7.0$ (Basic) | $K_2CO_3$, $CH_3COONa$, $NaCN$ |
| Weak | Weak | Both ions react with water. | Depends on $K_a$ and $K_b$ | $NH_4CN$, $NH_4NO_2$ |
4. Conjugate Theory Rules
- Anions of weak acids (e.g., $CH_3COO^-$, $CN^-$) are strong conjugate bases and will react with water to produce $OH^-$ ions.
- Cations of weak bases (e.g., $NH_4^+$, $Cu^{2+}$) are strong conjugate acids and will react with water to produce $H^+$ ions.
- Ions from strong acids/bases (e.g., $Na^+$, $Cl^-$) are too weak to react with water.
Questions and Answers
Q1: Why does an aqueous solution of $NH_4Cl$ turn blue litmus red?
A: $NH_4Cl$ is a salt of a strong acid ($HCl$) and a weak base ($NH_4OH$). The $NH_4^+$ cation undergoes hydrolysis:
$$NH_4^+_{(aq)} + H_2O_{(l)} \rightleftharpoons NH_4OH_{(aq)} + H^+_{(aq)}$$
The production of $H^+$ ions makes the solution acidic, causing the blue litmus to turn red.
Q2: Explain why $NaCl$ has no action on litmus paper.
A: $NaCl$ is derived from a strong acid ($HCl$) and a strong base ($NaOH$). Both $Na^+$ and $Cl^-$ ions are very weak conjugates and do not react with water. Since no excess $H^+$ or $OH^-$ ions are produced, the solution remains neutral ($pH = 7.0$).
Q3: Distinguish between Hydration and Hydrolysis.
A: In hydrolysis, the chemical bond of the water molecule ($H-OH$) is broken to form new compounds. In hydration, water molecules surround the solute particles via ion-dipole or dipole-dipole forces without breaking the water molecule's internal bonds.
Q4: What determines the pH of a salt solution formed from a weak acid and a weak base?
A: The pH depends on the relative values of the dissociation constants ($K_a$ and $K_b$) of the cation and anion. The solution could be slightly acidic, slightly basic, or neutral depending on which ion reacts more extensively with water.
Q5: Write the hydrolysis reaction for the carbonate ion ($CO_3^{2-}$).
A: The reaction is:
$$CO_3^{2-}_{(aq)} + 2H_2O_{(l)} \rightleftharpoons H_2CO_3_{(aq)} + 2OH^-_{(aq)}$$
Because $OH^-$ ions are produced, the resulting solution is basic ($pH > 7$).
9.10 Solubility Product and Precipitation Reactions
When a sparingly soluble ionic compound is mixed with water, a dynamic equilibrium is established between the undissolved solid and the ions in a saturated solution.
9.10.1 Solubility Product ($K_{sp}$)
The Solubility Product Constant ($K_{sp}$) is the product of the equilibrium concentrations of the ions, each raised to a power equal to its coefficient in the balanced chemical equation.
General Expression: For a compound $A_mB_n$, the equilibrium is:$$A_mB_{n(s)} \rightleftharpoons mA^{n+}_{(aq)} + nB^{m-}_{(aq)}$$The $K_{sp}$ expression is:$$K_{sp} = [A^{n+}]^m [B^{m-}]^n$$
9.10.2 Precipitation Reactions
A precipitation reaction occurs when two solutions are mixed and form an insoluble solid. To predict if a precipitate will form, we calculate the Ion Product ($Q'$) using initial concentrations.
| Condition | Result |
|---|---|
| $Q' > K_{sp}$ | Precipitation occurs (Supersaturated) |
| $Q' < K_{sp}$ | No precipitation occurs (Unsaturated) |
Concept Assessment Exercise 9.5: Questions & Answers
i. Write $K_{sp}$ expressions for:
- a) Iron (II) Hydroxide: $Fe(OH)_2 \rightleftharpoons Fe^{2+} + 2OH^-$ Answer: $K_{sp} = [Fe^{2+}][OH^-]^2$
- b) Calcium Sulphate: $CaSO_4 \rightleftharpoons Ca^{2+} + SO_4^{2-}$ Answer: $K_{sp} = [Ca^{2+}][SO_4^{2-}]$
ii. Lead (II) Sulphate ($PbSO_4$) has a $K_{sp} = 1.96 \times 10^{-8}$ at $25^\circ C$. What is its solubility?
Solution: Let solubility be $s$. $PbSO_{4(s)} \rightleftharpoons Pb^{2+}_{(aq)} + SO_{4(aq)}^{2-}$ $K_{sp} = [Pb^{2+}][SO_4^{2-}] = (s)(s) = s^2$ $s = \sqrt{K_{sp}} = \sqrt{1.96 \times 10^{-8}}$ Answer: $1.4 \times 10^{-4} \text{ M}$
iii. In Indus river water, $[Ca^{2+}] = 1.0 \times 10^{-9} \text{ M}$ and $[PO_4^{3-}] = 1.0 \times 10^{-9} \text{ M}$. Will $Ca_3(PO_4)_2$ precipitate if $K_{sp} = 1.2 \times 10^{-29}$?
Solution: $Q' = [Ca^{2+}]^3 [PO_4^{3-}]^2$ $Q' = (1.0 \times 10^{-9})^3 \times (1.0 \times 10^{-9})^2$ $Q' = 1.0 \times 10^{-45}$ Compare $Q'$ and $K_{sp}$: $1.0 \times 10^{-45} < 1.2 \times 10^{-29}$ Answer: Since $Q' < K_{sp}$, No precipitation will occur.
Notes: Common Ion Effect
The Common Ion Effect is a chemical phenomenon that occurs when a soluble compound is added to a solution that already contains one of its constituent ions. This addition shifts the chemical equilibrium according to Le Chatelier's Principle.
1. Definition
The phenomenon in which the degree of ionization or solubility of an electrolyte is reduced by the addition of a highly soluble electrolyte containing a common ion is called the common ion effect.
2. Mechanism
- When a strong electrolyte (which dissociates completely) is added to a solution of a weak electrolyte (which dissociates partially), they share a common ion.
- The concentration of that specific ion increases significantly in the solution.
- To counteract this stress, the equilibrium of the weak electrolyte shifts toward the left (the undissociated side).
- This results in decreased ionization of the weak electrolyte or the precipitation of a less soluble salt.
3. Key Examples from Text
| System | Common Ion | Resulting Effect |
|---|---|---|
| $HF$ (Weak Acid) + $NaF$ (Salt) | $F^{-}$ | Dissociation of $HF$ is reduced. |
| $KClO_{4}$ (Saturated) + $KCl$ | $K^{+}$ | $KClO_{4}$ precipitates out of solution. |
| $NaCl$ (Brine) + $HCl$ (Gas) | $Cl^{-}$ | Pure $NaCl$ precipitates. |
Concept Assessment Exercise 9.6: Questions & Answers
Question i
Ammonium Chloride, $NH_{4}Cl$, is a water-soluble salt. What will happen if this salt is added to a solution containing ammonium hydroxide?
$$NH_{4}OH_{(aq)} \rightleftharpoons NH_{4}^{+}{}_{(aq)} + OH^{-}{}_{(aq)}$$
Answer: Ammonium hydroxide ($NH_{4}OH$) is a weak base that ionizes slightly. $NH_{4}Cl$ is a strong electrolyte that dissociates completely into $NH_{4}^{+}$ and $Cl^{-}$ ions. The addition of $NH_{4}Cl$ increases the concentration of the common ion, $NH_{4}^{+}$. According to Le Chatelier's Principle, the equilibrium will shift to the left to consume the excess ions, thereby suppressing the ionization of $NH_{4}OH$.
Question ii
Carbonic acid is a weak acid. It ionizes in water as follows:
$$H_{2}CO_{3(aq)} \rightleftharpoons 2H^{+}{}_{(aq)} + CO_{3}^{2-}{}_{(aq)}$$
What will happen if a strong electrolyte such as $Na_{2}CO_{3}$ is added to a solution containing carbonic acid?
Answer: $Na_{2}CO_{3}$ dissociates completely to provide a high concentration of $CO_{3}^{2-}$ ions. Because $CO_{3}^{2-}$ is common to the ionization of carbonic acid, the increase in its concentration forces the equilibrium of the $H_{2}CO_{3}$ reaction to shift to the left. Consequently, the dissociation of carbonic acid is reduced, and the concentration of undissociated $H_{2}CO_{3}$ increases.