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A Level H1 Chemistry Acids Bases Salts Quiz

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Questions

A-Level Chemistry H1 Quiz - Acids Bases Salts

Name: ________________________
Class: ________________________
Date: ________________________
Score: ________ / 50

Duration: 60 Minutes
Total Marks: 50
Instructions: Answer all questions. Show all working for calculations. Use the provided data booklet for $K_w$ and $K_a$ values where necessary.

Section A: Foundational Concepts (Questions 1–5)

Define the term weak acid and provide a balanced chemical equation, including state symbols, for the dissociation of ethanoic acid in water. [2]

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Distinguish between a strong acid and a concentrated acid. [2]

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Identify the Period 3 element that forms a sparingly soluble amphoteric oxide. [1]

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State the Brønsted-Lowry definition of a base. [1]

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Explain why the pH of a $0.10\text{ mol dm}^{-3}$ solution of $\text{HCl}$ is lower than the pH of a $0.10\text{ mol dm}^{-3}$ solution of $\text{CH}_3\text{COOH}$ . [2]

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Section B: Quantitative Analysis & Calculations (Questions 6–15)

Calculate the pH of a $0.050\text{ mol dm}^{-3}$ solution of nitric acid ( $\text{HNO}_3$ ). [1]

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A $25.0\text{ cm}^3$ sample of a weak monoprotic acid $\text{HA}$ was titrated against $0.100\text{ mol dm}^{-3}$ $\text{NaOH}$ . The average titre volume was $18.20\text{ cm}^3$ . Calculate the concentration of the acid. [3]

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For the acid in Question 7, the pH of the solution was found to be 3.20. Calculate the acid dissociation constant, $K_a$ , for $\text{HA}$ . [3]

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Write the expression for the acid dissociation constant, $K_a$ , for the first dissociation of carbonic acid ( $\text{H}_2\text{CO}_3$ ). [1]

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Calculate the pH of a $0.010\text{ mol dm}^{-3}$ solution of ammonia ( $\text{NH}_3$ ), given that $K_b = 1.8 \times 10^{-5}\text{ mol dm}^{-3}$ . [3]

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A buffer solution is prepared by mixing $0.15\text{ mol dm}^{-3}$ propanoic acid ( $\text{CH}_3\text{CH}_2\text{COOH}$ ) and $0.25\text{ mol dm}^{-3}$ sodium propanoate. Given $pK_a = 4.87$ , calculate the pH of this buffer. [3]

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How would the pH of the buffer in Question 11 change if a small amount of $0.1\text{ mol dm}^{-3}$ $\text{HCl}$ is added? Explain your answer using a chemical equation. [3]

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Calculate the mass of $\text{Na}_2\text{CO}_3$ required to prepare $250\text{ cm}^3$ of a $0.020\text{ mol dm}^{-3}$ solution. [2]

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A solution of a weak acid $\text{HA}$ has a $K_a$ of $1.0 \times 10^{-5}$ . If the concentration of the acid is doubled, what happens to the pH? Justify your answer. [2]

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Determine the pH of a solution formed by mixing $50\text{ cm}^3$ of $0.10\text{ mol dm}^{-3}$ $\text{HCl}$ and $50\text{ cm}^3$ of $0.10\text{ mol dm}^{-3}$ $\text{NaOH}$ . [2]

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Section C: Application & Reasoning (Questions 16–20)

In a fermentation tank, calcium hydroxide is added to prevent the buildup of lactic acid. Explain why excessive acidity (low pH) would reduce the effectiveness of the enzymes involved in fermentation. [2]

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Compare the solubility of $\text{Al}_2\text{O}_3$ in $\text{HCl}(\text{aq})$ and $\text{NaOH}(\text{aq})$ . Provide one balanced equation to support your answer. [2]

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A salt is formed by the reaction between a strong acid and a weak base. Predict whether the resulting solution will be acidic, basic, or neutral. Explain your reasoning. [2]

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Describe the effect of increasing the temperature on the value of $K_w$ for water. How does this affect the neutrality of water at $60^\circ\text{C}$ ? [2]

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An unknown salt $X$ is found to be insoluble in water but dissolves in both concentrated $\text{HCl}$ and concentrated $\text{NaOH}$ . Identify the nature of the oxide/salt $X$ and give an example of such a compound. [2]

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Answers

A-Level Chemistry H1 Quiz - Acids Bases Salts (Answer Key)

Section A: Foundational Concepts

Definition: A weak acid is an acid that only partially dissociates/ionizes in aqueous solution. [1] Equation: $\text{CH}_3\text{COOH}(\text{aq}) \rightleftharpoons \text{CH}_3\text{COO}^-(\text{aq}) + \text{H}^+(\text{aq})$ (Must have equilibrium arrow and state symbols). [1]
Strong acid: Completely dissociates in water to produce $\text{H}^+$ ions. [1] Concentrated acid: Has a high molarity/concentration of solute per unit volume of solvent. [1]
Aluminium (Al). [1]
Brønsted-Lowry Base: A species (molecule or ion) that acts as a proton ( $\text{H}^+$ ) acceptor. [1]
$\text{HCl}$ is a strong acid and dissociates completely, resulting in a higher concentration of $\text{H}^+$ ions compared to $\text{CH}_3\text{COOH}$ , which is a weak acid and only partially dissociates. Since $\text{pH} = -\log[\text{H}^+]$ , a higher $[\text{H}^+]$ leads to a lower pH. [2]

Section B: Quantitative Analysis & Calculations

$[\text{H}^+] = 0.050\text{ mol dm}^{-3}$ ; $\text{pH} = -\log(0.050) = 1.30$ . [1]
$\text{n}(\text{NaOH}) = 0.100 \times (18.20/1000) = 0.00182\text{ mol}$ . [1] $\text{n}(\text{HA}) = 0.00182\text{ mol}$ (1:1 ratio). [1] $\text{Conc}(\text{HA}) = 0.00182 / (25.0/1000) = 0.0728\text{ mol dm}^{-3}$ . [1]
$\text{pH} = 3.20 \implies [\text{H}^+] = 10^{-3.20} = 6.31 \times 10^{-4}\text{ mol dm}^{-3}$ . [1] $K_a = [\text{H}^+]^2 / [\text{HA}] = (6.31 \times 10^{-4})^2 / 0.0728$ . [1] $K_a = 5.47 \times 10^{-6}\text{ mol dm}^{-3}$ . [1]
$K_a = \frac{[\text{HCO}_3^-][\text{H}^+]}{[\text{H}_2\text{CO}_3]}$ [1]
$[\text{OH}^-] = \sqrt{K_b \times \text{conc}} = \sqrt{1.8 \times 10^{-5} \times 0.010} = 4.24 \times 10^{-4}\text{ mol dm}^{-3}$ . [1] $\text{pOH} = -\log(4.24 \times 10^{-4}) = 3.37$ . [1] $\text{pH} = 14 - 3.37 = 10.63$ . [1]
$\text{pH} = pK_a + \log([\text{salt}]/[\text{acid}]) = 4.87 + \log(0.25/0.15)$ . [1] $\text{pH} = 4.87 + 0.22 = 5.09$ . [2]
The pH will decrease slightly. [1] The added $\text{H}^+$ reacts with the conjugate base: $\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \to \text{CH}_3\text{CH}_2\text{COOH}$ . [2]
$\text{n} = 0.020 \times 0.250 = 0.005\text{ mol}$ . [1] $\text{Mass} = 0.005 \times 106.0 = 0.53\text{ g}$ . [1]
The pH will decrease. [1] Since $[\text{H}^+] \approx \sqrt{K_a \times [\text{HA}]}$ , increasing the concentration of the weak acid increases the concentration of $\text{H}^+$ ions produced. [1]
$\text{n}(\text{HCl}) = 0.005\text{ mol}$ ; $\text{n}(\text{NaOH}) = 0.005\text{ mol}$ . [1] They neutralize completely to form $\text{NaCl}$ and $\text{H}_2\text{O}$ . The solution is neutral, $\text{pH} = 7.0$ . [1]

Section C: Application & Reasoning

Low pH (high acidity) denatures the enzyme by disrupting the ionic and hydrogen bonds in its tertiary structure. [1] This changes the shape of the active site, preventing the substrate from binding. [1]
$\text{Al}_2\text{O}_3$ is amphoteric; it is soluble in both. [1] Equation: $\text{Al}_2\text{O}_3(\text{s}) + 6\text{HCl}(\text{aq}) \to 2\text{AlCl}_3(\text{aq}) + 3\text{H}_2\text{O}(\text{l})$ OR $\text{Al}_2\text{O}_3(\text{s}) + 2\text{NaOH}(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \to 2\text{Na}[\text{Al}(\text{OH})_4](\text{aq})$ . [1]
Acidic. [1] The conjugate acid of the weak base will undergo hydrolysis in water, releasing $\text{H}^+$ ions. [1]
$K_w$ increases as temperature increases (dissociation of water is endothermic). [1] The pH of neutral water will be lower than 7.0 (e.g., $\sim 6.5$ ), but it remains "neutral" as $[\text{H}^+] = [\text{OH}^-]$ . [1]
Amphoteric. [1] Example: Aluminium oxide ( $\text{Al}_2\text{O}_3$ ) or Zinc oxide ( $\text{ZnO}$ ). [1]