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A Level H2 Chemistry Acids Bases Salts Quiz

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Questions

A-Level Chemistry H2 Quiz - Acids Bases Salts

Name: ____________________ Class: ____________________ Date: ____________________ Score: ________ / 55

Duration: 90 Minutes
Total Marks: 55
Instructions: Answer all questions. Use the Data Booklet where necessary. Show all working for calculations.

Section A: Qualitative Analysis & Gas Tests (Questions 1-6)

Complete the following table for the identification of gases. [4]

Gas	Test and Result
Ammonia, $\text{NH}_3$
Carbon dioxide, $\text{CO}_2$

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A student is testing for the presence of $\text{SO}_2$ and $\text{CO}_2$ . Both gases can react with limewater. Suggest a specific test to distinguish between $\text{SO}_2$ and $\text{CO}_2$ . [2]

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Complete the table for the reactions of the following aqueous cations with $\text{NaOH(aq)}$ and $\text{NH}_3\text{(aq)}$ . [4]

Cation	Reaction with $\text{NaOH(aq)}$	Reaction with $\text{NH}_3\text{(aq)}$
$\text{Al}^{3+}(\text{aq})$
$\text{Cu}^{2+}(\text{aq})$

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State the observation when excess aqueous ammonia is added to a solution containing $\text{Zn}^{2+}(\text{aq})$ ions. [1]

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Write an ionic equation for the reaction of $\text{Al}_2\text{O}_3(\text{s})$ with hot aqueous sodium hydroxide. [2]

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Explain why $\text{Pb}(\text{OH})_2$ is described as amphoteric. [2]

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Section B: Titrations & Quantitative Analysis (Questions 7-14)

A student performs three titrations to determine the concentration of a weak acid $\text{HA}$ . The volumes of $\text{FA}$ used are $24.10\text{ cm}^3$ , $23.90\text{ cm}^3$ , and $23.95\text{ cm}^3$ . (a) Identify the concordant results. [1] (b) Calculate the suitable volume of $\text{FA}$ to be used in calculations. [2]

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Calculate the number of moles of $\text{HCl}$ present in $25.0\text{ cm}^3$ of $0.100\text{ mol dm}^{-3}$ solution. [2]

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A titration requires $22.50\text{ cm}^3$ of $0.150\text{ mol dm}^{-3}$ $\text{NaOH}$ to neutralize $25.0\text{ cm}^3$ of $\text{H}_2\text{SO}_4$ . Calculate the concentration of the sulfuric acid. [3]

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In a calorimetry experiment, $50.0\text{ cm}^3$ of $1.0\text{ mol dm}^{-3}$ $\text{NaOH}$ is mixed with $50.0\text{ cm}^3$ of $1.0\text{ mol dm}^{-3}$ $\text{HCl}$ . (a) Calculate the moles of $\text{NaOH}$ added. [1] (b) Identify the limiting reagent if $50.0\text{ cm}^3$ of $0.8\text{ mol dm}^{-3}$ $\text{HCl}$ was used instead. Show your working. [3]

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Define the term "buffer solution". [2]

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Calculate the pH of a $0.10\text{ mol dm}^{-3}$ solution of $\text{CH}_3\text{COOH}$ given that $K_a = 1.8 \times 10^{-5}\text{ mol dm}^{-3}$ . [3]

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A buffer solution is prepared by mixing $0.20\text{ mol}$ of $\text{CH}_3\text{COOH}$ and $0.20\text{ mol}$ of $\text{CH}_3\text{COONa}$ in $1.0\text{ dm}^3$ of solution. Calculate the pH of this buffer. [3]

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Explain the effect on the pH of the buffer in Question 13 when a small amount of $\text{HCl}$ is added. [2]

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Section C: Advanced Equilibria & Salts (Questions 15-20)

Write the expression for the solubility product, $K_{sp}$ , of $\text{CaF}_2(\text{s})$ . [2]

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The $K_{sp}$ of $\text{AgCl}(\text{s})$ is $1.8 \times 10^{-10}\text{ mol}^2\text{ dm}^{-6}$ . Calculate the solubility of $\text{AgCl}$ in pure water. [3]

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Explain why the solubility of $\text{AgCl}(\text{s})$ increases when added to a solution of $\text{NH}_3(\text{aq})$ . [3]

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Predict the pH of a $0.1\text{ mol dm}^{-3}$ solution of $\text{AlCl}_3(\text{aq})$ . Justify your answer with an equation. [3]

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Compare the strength of $\text{HClO}_4$ and $\text{HClO}_3$ as acids. Explain your reasoning. [3]

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A salt is formed by the reaction of a strong acid and a weak base. Predict whether the resulting aqueous solution of the salt will be acidic, basic, or neutral. Explain your answer. [3]

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Answers

Answer Key - A-Level Chemistry H2 Quiz (Acids Bases Salts)

Ammonia: Turns damp red litmus paper blue. [1] Carbon dioxide: Gives a white precipitate with limewater; precipitate dissolves in excess $\text{CO}_2$ . [1] (Note: 2 marks for each complete entry) [4]
$\text{SO}_2$ bleaches damp litmus paper (red/blue turns white), whereas $\text{CO}_2$ does not. [2]
$\text{Al}^{3+}$ : $\text{NaOH}$ : White ppt, soluble in excess. [1] $\text{NH}_3$ : White ppt, insoluble in excess. [1] $\text{Cu}^{2+}$ : $\text{NaOH}$ : Blue ppt, insoluble in excess. [1] $\text{NH}_3$ : Blue ppt, soluble in excess (forming deep blue solution). [1] [4]
White precipitate dissolves in excess to form a colorless solution. [1]
$\text{Al}_2\text{O}_3(\text{s}) + 2\text{OH}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow 2[\text{Al}(\text{OH})_4]^-(\text{aq})$ [2]
It reacts with both acids (e.g., $\text{HCl}$ ) and strong bases (e.g., $\text{NaOH}$ ) to form soluble salts. [2]
(a) $23.90\text{ cm}^3$ and $23.95\text{ cm}^3$ . [1] (b) Mean = $(23.90 + 23.95) / 2 = 23.93\text{ cm}^3$ (or $23.92$ depending on rounding). [2]
$n = c \times V = 0.100 \times (25.0/1000) = 2.50 \times 10^{-3}\text{ mol}$ . [2]
$\text{n}(\text{NaOH}) = 0.150 \times 0.02250 = 3.375 \times 10^{-3}\text{ mol}$ . $\text{n}(\text{H}_2\text{SO}_4) = 3.375 \times 10^{-3} / 2 = 1.6875 \times 10^{-3}\text{ mol}$ . $c = 1.6875 \times 10^{-3} / 0.0250 = 0.0675\text{ mol dm}^{-3}$ . [3]
(a) $n = 1.0 \times 0.050 = 0.050\text{ mol}$ . [1] (b) $\text{n}(\text{NaOH}) = 0.050\text{ mol}$ . $\text{n}(\text{HCl}) = 0.8 \times 0.050 = 0.040\text{ mol}$ . Since $0.040 < 0.050$ and ratio is 1:1, $\text{HCl}$ is the limiting reagent. [3]
A solution that resists significant changes in pH when small amounts of acid or base are added. [2]
$[\text{H}^+] = \sqrt{K_a \times c} = \sqrt{1.8 \times 10^{-5} \times 0.10} = 1.34 \times 10^{-3}\text{ mol dm}^{-3}$ . $\text{pH} = -\log(1.34 \times 10^{-3}) = 2.87$ . [3]
$\text{pH} = \text{p}K_a + \log([\text{salt}]/[\text{acid}])$ . $\text{p}K_a = -\log(1.8 \times 10^{-5}) = 4.74$ . $\text{pH} = 4.74 + \log(0.2/0.2) = 4.74$ . [3]
$\text{HCl}$ reacts with the conjugate base ( $\text{CH}_3\text{COO}^-$ ) to form $\text{CH}_3\text{COOH}$ . $\text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH}$ . The ratio of $[\text{salt}]/[\text{acid}]$ changes only slightly, so pH remains nearly constant. [2]
$K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2$ [2]
$s^2 = 1.8 \times 10^{-10} \rightarrow s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}\text{ mol dm}^{-3}$ . [3]
$\text{Ag}^+$ reacts with $\text{NH}_3$ to form the stable complex $[\text{Ag}(\text{NH}_3)_2]^+$ . [1] This reduces the concentration of free $\text{Ag}^+$ ions in solution. [1] According to Le Chatelier's principle, the equilibrium $\text{AgCl}(\text{s}) \rightleftharpoons \text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq})$ shifts to the right. [1] [3]
Acidic. [1] $\text{Al}^{3+}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons [\text{Al}(\text{OH})(\text{H}_2\text{O})_5]^{2+}(\text{aq}) + \text{H}^+(\text{aq})$ [2]
$\text{HClO}_4$ is stronger. [1] The perchlorate ion $\text{ClO}_4^-$ is more stable/weaker conjugate base than $\text{ClO}_3^-$ . [1] Due to higher oxidation state of $\text{Cl}$ and greater electron-withdrawing effect, polarising the $\text{O-H}$ bond more. [1] [3]
Acidic. [1] The salt contains the conjugate acid of a weak base. [1] This conjugate acid partially dissociates in water to release $\text{H}^+$ ions (hydrolysis). [1] [3]