i. The value of the ionic product of water
Answer: b) depends on temperature
Explanation: The ionic product of water (Kw) is temperature-dependent. It increases with increasing temperature because the autoionization of water is endothermic.
Part A: Multiple Choice Questions (MCQs)
ii. A base when dissolved in water yields a solution with a hydroxyl ion concentration of 0.05 mol dm⁻³. The solution is
Answer: a) Basic
Explanation: A solution with a high concentration of hydroxide ions (OH⁻) relative to hydrogen ions (H⁺) is basic. The pOH is -
log(0.05) [source symbol unavailable] 1.3, so the pH is 14 - 1.3 = 12.7, which is strongly basic.
iii. pH scale was introduced by:
Answer: b) Sorensen
Explanation: The pH scale was introduced by the Danish chemist Søren Peder Lauritz Sørensen in 1909.
iv. pH of solution is defined by expression
Answer: b) log(1/[H⁺])
Explanation: pH is defined as the negative logarithm of the hydrogen ion concentration: pH = -log[H⁺] = log(1/[H⁺]).
v. The pH of a 10⁻³ M HCl solution at 25°C if it is diluted 1000 times, will be
Answer: c) 5.98
Explanation: After dilution, [H⁺] from HCl = 10⁻⁶ M. This is close to the [H⁺] from water (10⁻⁷ M), so both must be considered. Total [H⁺]
= 10⁻⁶ + 10⁻⁷ = 1.1 × 10⁻⁶ M. pH = -log(1.1 × 10⁻⁶) = 5.96 [source symbol unavailable] 5.98.
vi. How many dm³ of water must be added to 1 dm³ an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?
Answer: d) 9.0 dm³
Explanation: Initial [H⁺] = 10⁻¹ = 0.1 M. Final [H⁺] = 10⁻² = 0.01 M. M₁V₁ = M₂V₂ [source symbol unavailable] 0.1 × 1 = 0.01 × V₂ [source symbol unavailable] V₂ = 10 dm³. Water to add = 10 - 1
= 9 dm³.
vii. What is the approximate pH of a 1 × 10⁻⁷ M NaOH solution?
Answer: c) 7
Explanation: At this very low concentration, the OH⁻ from water (10⁻⁷ M) cannot be ignored. Total [OH⁻] [source symbol unavailable] 2 × 10⁻⁷ M. pOH [source symbol unavailable] 6.7, so pH [source symbol unavailable] 7.3, which is approximately 7.
viii. Calculate the pOH of a solution at 25°C that contains 1 × 10⁻¹⁰ M of hydronium ions, i.e. H₃O⁺.
Answer: a) 4.0
Explanation: pH = -log(1 × 10⁻¹⁰) = 10. pOH = 14 - pH = 14 - 10 = 4.
ix. The pH value of a 10 M solution of HCl is
Answer: a) less than 0
Explanation: pH = -log(10) = -1. A negative pH is possible for very concentrated strong acids.
x. Which of the following has the highest pH?
Answer: d) M/4 Ca(OH)₂
Explanation: Ca(OH)₂ is a strong base that provides 2 OH⁻ ions per formula unit. So [OH⁻] = 2 × M/4 = M/2. This is the highest OH⁻ concentration, hence the highest pH.
xi. Which of the following statements are correct?
(i) Kw = [H⁺][OH⁻] = 10⁻¹⁴ mol²dm⁻⁶ at 298K
(ii) At 298K [H⁺] = [OH⁻] = 10⁻⁷
(iii) Kw does not depend upon temperature
(iv) Molarity of pure water = 55.55 M
Answer: b) (i), (ii) and (iv)
Explanation: Statement (iii) is incorrect because Kw is temperature-dependent. Statements (i), (ii), and (iv) are all correct.
Part B: Short Answer Questions
i. Calculate H⁺ ion concentration of a solution prepared by dissolving 4 g of NaOH (Atomic weight of Na = 23 amu) in 1000 cm³ of solution?
Answer:
Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol.
Moles of NaOH = 4 g / 40 g/mol = 0.1 mol.
Volume = 1000 cm³ = 1 dm³.
[OH⁻] = 0.1 M.
pOH = -log(0.1) = 1.
pH = 14 - 1 = 13.
[H⁺] = 10⁻¹³ M.
ii. Calculate the pH of 0.005 molar solution of H₂SO₄.
Answer:
H₂SO₄ is a strong diprotic acid. [H⁺] = 2 × 0.005 M = 0.01 M.
pH = -log(0.01) = 2.
iii. Calculate the pH of the following compounds:
(i) 10⁻⁴ M KOH
Answer: [OH⁻] = 10⁻⁴ M. pOH = 4. pH = 14 - 4 = 10.
(ii) 10⁻⁵ M HCl
Answer: [H⁺] = 10⁻⁵ M. pH = 5.
(iii) 10⁻⁵ M KOH
Answer: [OH⁻] = 10⁻⁵ M. pOH = 5. pH = 14 - 5 = 9.
(iv) 10⁻⁴ M HCl
Answer: [H⁺] = 10⁻⁴ M. pH = 4.
iv. 100 cm³ of 0.04 M HCl aqueous solution is mixed with 100 cm³ of 0.02 M NaOH solution. Calculate the pH of the resulting solution.
Answer:
Moles of HCl = 0.04 × 0.1 = 0.004 mol.
Moles of NaOH = 0.02 × 0.1 = 0.002 mol.
Excess HCl = 0.004 - 0.002 = 0.002 mol.
Total volume = 200 cm³ = 0.2 dm³.
[H⁺] = 0.0020.2 = 0.01 M.
pH = -log(0.01) = 2.
v. Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. What will be the H⁺ ion concentration in the mixture?
Answer:
[H⁺] from pH 3 = 10⁻³ M.
[H⁺] from pH 4 = 10⁻⁴ M.
[H⁺] from pH 5 = 10⁻⁵ M.
Let volume of each = V. Total volume = 3V.
Total moles H⁺ = V(10⁻³ + 10⁻⁴ + 10⁻⁵) = V(0.001 + 0.0001 + 0.00001) = V(0.00111).
[H⁺] = 0.00111 V / 3V = 3.7 × 10⁻⁴ M.
vi. A 20.0 cm³ sample of 0.200 mol dm⁻³ NH₃(aq) was titrated with 0.100 mol dm⁻³ HCl. On the following axes, sketch how the pH changes during this titration. Mark clearly where the end point occurs.
Answer: (Diagram Description)
Initial pH: Weak base, pH ~11. Buffer Region: Gradual decrease in pH as HCl is added, forming NH₄⁺/NH₃ buffer.
Equivalence Point: At 40 cm³ HCl added (moles NH₃ = 0.004, moles HCl = 0.004), pH < 7 (acidic due to NH₄⁺). The curve has a steep drop. After Equivalence: pH approaches that of excess HCl. Endpoint: The steep part of the curve around pH 5-6.
Part C: Long Answer Questions
i. Explain the concept of pH and its significance in acid-base chemistry. How does the pH scale relate to the concentration of hydrogen ions in a solution?
Answer: The concept of pH was introduced by Sørensen to express the acidity or alkalinity of a solution in a convenient way. pH is defined as
the negative logarithm (base 10) of the hydrogen ion concentration: pH = -log[H⁺].
Significance:
Acidity/Alkalinity: A low pH (0-6.9) indicates acidity, pH 7 is neutral, and a high pH (7.1-14) indicates alkalinity. Biological Importance: Enzymes and biological processes function optimally within narrow pH ranges (e.g., blood pH ~7.4, stomach pH ~2). Industrial Applications: pH control is crucial in agriculture, food processing, water treatment, and chemical manufacturing.
Relationship to [H⁺]:
The pH scale is logarithmic. A change of one pH unit corresponds to a tenfold change in [H⁺].
For example, a solution with pH 3 has [H⁺] = 10⁻³ M, and a solution with pH 4 has [H⁺] = 10⁻⁴ M. The pH scale typically ranges from 0 to 14 at 25°C, corresponding to [H⁺] from 10⁰ M to 10⁻¹⁴ M.
In summary, pH provides a simple numerical scale to express the concentration of hydrogen ions, which is fundamental to understanding acid-base chemistry and its applications.
ii. Compare and contrast the titration curves for a strong acid with a strong base, a weak acid with a strong base, a strong acid with a weak base, and a weak acid with a weak base.
Answer: Titration curves show the change in pH as a base is added to an acid (or vice versa). The shape of the curve depends on the strengths of the acid and base.
1. Strong Acid - Strong Base:
Initial pH: Low (~1).
Equivalence Point: pH = 7. Curve: Sharp, steep rise at the equivalence point. Indicators like phenolphthalein or methyl orange can be used.
2. Weak Acid - Strong Base:
Initial pH: Higher than strong acid (~3). Buffer Region: Gradual rise as a buffer forms. Equivalence Point: pH > 7 (basic due to conjugate base). Curve: Less steep than strong-strong. Phenolphthalein is a suitable indicator.
3. Strong Acid - Weak Base:
Initial pH: Low (~1). Equivalence Point: pH < 7 (acidic due to conjugate acid). Curve: Steep drop at equivalence. Methyl orange is a suitable indicator.
4. Weak Acid - Weak Base:
Initial pH: Higher than strong acid. Equivalence Point: pH depends on relative strengths; no sharp change. Curve: Very gradual, no sharp endpoint. Titration is not very accurate.
In summary, the titration curve shape and the pH at the equivalence point are determined by the acid and base strengths.
iii. A 50 cm³ solution of 0.1 M acetic acid is titrated with 0.1 M sodium hydroxide. The Ka of acetic acid at 25 °C is 1.8 × 10⁻⁵ mol dm⁻³. Calculate the pH of the solution at the equivalence point and explain the shape of the titration curve for this reaction.
Answer:
At Equivalence Point: Moles of acid = moles of base. 50 cm³ of 0.1 M CH₃COOH requires 50 cm³ of 0.1 M NaOH. Total
volume = 100 cm³ = 0.1 dm³.
Concentration of Salt (CH₃COONa): Moles = 0.1 × 0.05 = 0.005 mol. [CH₃COONa] = 0.0050.1 = 0.05 M. Hydrolysis of Salt: CH₃COO⁻ + H₂O [source symbol unavailable] CH₃COOH + OH⁻
Kb for Acetate: Kb = Kw / Ka = 1 × 10⁻¹⁴1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰.
Calculate [OH⁻]: Kb = [OH⁻]² / [CH₃COO⁻] [source symbol unavailable] 5.56 × 10⁻¹⁰ = [OH⁻]² / 0.05.
[OH⁻]² = 2.78 × 10⁻¹¹ [source symbol unavailable] [OH⁻] = 5.27 × 10⁻⁶ M.
pOH: pOH = -log(5.27 × 10⁻⁶) = 5.28.
pH: pH = 14 - pOH = 14 - 5.28 = 8.72.
Shape of Titration Curve:
Initial pH: Weak acid (CH₃COOH), pH ~ 2.9. Buffer Region: Gradual rise in pH as CH₃COONa is formed. Equivalence Point: pH > 7 (8.72) due to the basic nature of the acetate ion. After Equivalence: pH rises rapidly due to excess NaOH.
In summary, the titration of a weak acid with a strong base has an equivalence point in the basic range.