This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Another one Bad — let's solve it. 7.1 7.1.1 The term for an ion of a salt reacting with water is hydrolysis. 7.1.2 The reaction CO_3^2-(aq) + H_2O(l) HCO_3^-(aq) + OH^-(aq) is a Lowry-Bronsted base reaction because the carbonate ion (CO_3^2-) accepts a proton (H^+) from water (H_2O), forming bicarbonate (HCO_3^-) and hydroxide (OH^-). According to the Lowry-Bronsted model, a base is a proton acceptor. 7.1.3 The conjugate base of water (H_2O) is OH^-. 7.1.4 The carbonate ion CO_3^2- is a weak base. For a weak base, its dissociation constant (K_b) must be greater than the autoionization constant of water (K_w = 1 × 10^-14 at 25 °C) but significantly less than 1. The correct option is K_b > 1 × 10^-14. 7.1.5 An ampholyte is a substance that can act as both an acid (proton donor) and a base (proton acceptor). 7.1.6 The conjugate acid of the hydrogen carbonate ion (HCO_3^-) is H_2CO_3. 7.1.7 The conjugate base of the hydrogen carbonate ion (HCO_3^-) is CO_3^2-. 7.2 7.2.1 A diprotic acid is an acid that can donate two protons (H^+ ions) per molecule in an aqueous solution. 7.2.2 Step 1: Calculate the hydrogen ion concentration from the given pH. The pH is given as 1.3. [H^+] = 10^-pH [H^+] = 10^-1.3 [H^+] = 0.0501187 mol·dm^-3 Step 2: Calculate the hydroxide ion concentration using the ion product of water (K_w). At 25 °C, K_w = [H^+][OH^-] = 1 × 10^-14. [OH^-] = (K_w)/([H)^+] [OH^-] = 1 × 10^-140.0501187 [OH^-] = 1.995 × 10^-13 mol·dm^-3 The concentration of the hydroxide ions is 1.995 × 10^-13 mol·dm^-3. 7.3 7.3.1 Step 1: Calculate the initial concentration of the diprotic acid. Since the acid is strong and diprotic, it fully dissociates, releasing two H^+ ions per molecule. [H^+] = 2 × C_acid, initial From 7.2.2, [H^+] = 0.0501187 mol·dm^-3. C_acid, initial = [H^+]2 C_acid, initial = (0.0501187)/(2) C_acid, initial = 0.02505935 mol·dm^-3 Step 2: Calculate the concentration of the dilute acid after dilution. The initial volume of acid is V_1 = 8 cm^3 = 0.008 dm^3. The final volume of the diluted solution is V_2 = 100 cm^3 = 0.100 dm^3. Using the dilution formula C_1V_1 = C_2V_2: (0.02505935 mol·dm^-3)(0.008 dm^3) = C_acid, dilute(0.100 dm^3) C_acid, dilute = ((0.02505935)(0.008))/(0.100) C_acid, dilute = 0.002004748 mol·dm^-3 Step 3: Calculate the concentration of the sodium hydroxide solution using titration data. The volume of dilute acid used is V_a = 25 cm^3 = 0.025 dm^3. The volume of sodium hydroxide used is V_b = 14.2 cm^3 = 0.0142 dm^3. The ratio of BASE : ACID is 2 : 1. This means n_b = 2 and n_a = 1 in the stoichiometry. Using the titration formula: (C_a V_a)/(n_a) = (C_b V_b)/(n_b) (0.002004748 mol·dm^-3)(0.025 dm^3)1 = C_b (0.0142 dm^3)2 C_b = ((0.002004748)(0.025)(2))/(0.0142) C_b = 0.0070589 mol·dm^-3 The concentration of the sodium hydroxide solution is 0.00706 mol·dm^-3. 7.3.2 Indicator B must be used in the titration. The titration involves a strong diprotic acid and a strong base (sodium hydroxide). The equivalence point for a strong acid-strong base titration occurs at a pH of approximately 7. Indicator B has a pH range of 6.8-7.4, which effectively covers the equivalence point of this titration. Send me the next one 📸