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Chemistry - Acids, Bases and Salts

Grade 10IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Acids are defined as proton (H+H^+) donors. In aqueous solutions, they dissociate to release hydrogen ions. Strong acids, such as HClHCl and H2SO4H_2SO_4, dissociate completely, whereas weak acids, like CH3COOHCH_3COOH, dissociate only partially.

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Bases are defined as proton (H+H^+) acceptors. Alkalis are bases that are soluble in water and release hydroxide ions (OH−OH^-). Examples include NaOHNaOH (strong) and NH3NH_3 (weak).

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The pHpH scale ranges from 00 to 1414 and measures the concentration of H+H^+ ions. A pH<7pH < 7 is acidic, pH=7pH = 7 is neutral (e.g., pure H2OH_2O), and pH>7pH > 7 is alkaline. Indicators like litmus (red in acid, blue in alkali), phenolphthalein (colorless in acid, pink in alkali), and methyl orange (red in acid, yellow in alkali) are used for testing.

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Oxides are classified into four categories: Basic oxides (metal oxides like CaOCaO), Acidic oxides (non-metal oxides like SO2SO_2), Amphoteric oxides (react with both acids and bases, e.g., Al2O3Al_2O_3 and ZnOZnO), and Neutral oxides (do not react with either, e.g., COCO and NONO).

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Salts are formed when the hydrogen ions of an acid are replaced by metal ions or ammonium ions (NH4+NH_4^+). The method of preparation depends on the solubility of the salt: Titration is used for soluble salts from alkalis, and Precipitation is used for making insoluble salts.

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Neutralization is an exothermic reaction between an acid and a base to produce a salt and water. The ionic equation for this process is always H+(aq)+OH−(aq)ightarrowH2O(l)H^+(aq) + OH^-(aq) ightarrow H_2O(l).

📐Formulae

Acid+Metal→Salt+H2(g)\text{Acid} + \text{Metal} \rightarrow \text{Salt} + H_2(g) (Example: 2HCl+Mg→MgCl2+H22HCl + Mg \rightarrow MgCl_2 + H_2)

Acid+Base→Salt+H2O(l)\text{Acid} + \text{Base} \rightarrow \text{Salt} + H_2O(l) (Example: H2SO4+CuO→CuSO4+H2OH_2SO_4 + CuO \rightarrow CuSO_4 + H_2O)

Acid+Metal Carbonate→Salt+H2O(l)+CO2(g)\text{Acid} + \text{Metal Carbonate} \rightarrow \text{Salt} + H_2O(l) + CO_2(g) (Example: 2HCl+CaCO3→CaCl2+H2O+CO22HCl + CaCO_3 \rightarrow CaCl_2 + H_2O + CO_2)

H+(aq)+OH−(aq)→H2O(l)H^+(aq) + OH^-(aq) \rightarrow H_2O(l) (General Ionic Equation for Neutralization)

n=c×Vn = c \times V (Where nn is moles, cc is concentration in mol/dm3mol/dm^3, and VV is volume in dm3dm^3)

💡Examples

Problem 1:

Predict the products and write the balanced chemical equation for the reaction between dilute Hydrochloric acid (HClHCl) and Calcium Carbonate (CaCO3CaCO_3).

Solution:

2HCl(aq)+CaCO3(s)→CaCl2(aq)+H2O(l)+CO2(g)2HCl(aq) + CaCO_3(s) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)

Explanation:

Acids reacting with carbonates always produce a salt, water, and carbon dioxide. In this case, the salt is Calcium Chloride (CaCl2CaCl_2).

Problem 2:

Explain how to prepare a pure, dry sample of the insoluble salt Barium Sulfate (BaSO4BaSO_4).

Solution:

  1. Mix aqueous solutions of Barium Chloride (BaCl2BaCl_2) and Sodium Sulfate (Na2SO4Na_2SO_4). 2. Filter the mixture to collect the precipitate (BaSO4BaSO_4). 3. Wash the precipitate with distilled water to remove impurities like NaClNaCl. 4. Dry the salt in a warm oven.

Explanation:

Since BaSO4BaSO_4 is an insoluble salt, the Precipitation method is used. The reaction is: BaCl2(aq)+Na2SO4(aq)→BaSO4(s)+2NaCl(aq)BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq). Filtering separates the solid salt from the soluble sodium chloride.

Problem 3:

A student titrates 25.0 cm325.0\ cm^3 of 0.10 mol/dm30.10\ mol/dm^3 NaOHNaOH with H2SO4H_2SO_4. If 12.5 cm312.5\ cm^3 of H2SO4H_2SO_4 is needed for neutralization, calculate the concentration of the acid.

Solution:

Reaction: 2NaOH+H2SO4→Na2SO4+2H2O2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O. Moles of NaOH=0.025×0.10=0.0025 molNaOH = 0.025 \times 0.10 = 0.0025\ mol. Molar ratio of NaOH:H2SO4NaOH:H_2SO_4 is 2:12:1. Moles of H2SO4=0.0025/2=0.00125 molH_2SO_4 = 0.0025 / 2 = 0.00125\ mol. Concentration of H2SO4=0.00125/0.0125=0.10 mol/dm3H_2SO_4 = 0.00125 / 0.0125 = 0.10\ mol/dm^3.

Explanation:

The stoichiometry of the balanced equation shows that 11 mole of H2SO4H_2SO_4 reacts with 22 moles of NaOHNaOH. This ratio is essential for calculating the concentration.

Acids, Bases and Salts - Revision Notes & Key Formulas | IGCSE Grade 10 Science