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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Here are the solutions to your questions: 7.1. 7.1.1. The ionization of water is ENDOTHERMIC. As the temperature increases from 25 °C to 50 °C, the value of K_w increases from 1.00 × 10^-14 to 5.48 × 10^-14. An increase in the equilibrium constant with increasing temperature indicates that the forward reaction (ionization) is favored by higher temperatures, which is characteristic of an endothermic process. 7.1.2. Step 1: Identify the K_w value at 50 °C. At 50 °C, K_w = 5.48 × 10^-14. Step 2: Use the relationship between K_w and [H_3O^+] for pure water. For pure water, [H_3O^+] = [OH^-]. K_w = [H_3O^+][OH^-] = [H_3O^+]^2 Step 3: Calculate [H_3O^+]. [H_3O^+] = sqrt(K_w) [H_3O^+] = sqrt(5.48 × 10^-14) [H_3O^+] = 2.34 × 10^-7 mol·dm^-3 The concentration of H_3O^+ ions in water at 50 °C is 2.34 × 10^-7 mol·dm^-3. 7.1.3. The pH of water will DECREASE. As the temperature increases, the value of K_w increases, which means the concentration of H_3O^+ ions increases. Since pH = -[H_3O^+], an increase in [H_3O^+] leads to a decrease in pH. 7.2. 7.2.1. Water acts as an acid (donates a proton to HCO_3^-): H_2O(l) + HCO_3^-(aq) OH^-(aq) + H_2CO_3(aq) 7.2.2. Water acts as a base (accepts a proton from HCO_3^-): H_2O(l) + HCO_3^-(aq) H_3O^+(aq) + CO_3^2-(aq) 7.3. 7.3.1. An electrolyte is a substance that, when dissolved in a solvent (typically water), produces ions and is therefore capable of conducting electricity. 7.3.2. The most effective electrolyte is H_2C_2O_4. The reason is that H_2C_2O_4 has the largest K_a value (6.0 × 10^-2) among the given acids. A larger K_a value indicates a stronger acid, meaning it ionizes to a greater extent in solution, producing a higher concentration of ions and thus making it a more effective electrolyte. Last free one today — make it count tomorrow, or type /upgrade for unlimited.