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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a) To write a balanced chemical equation for the reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl), we recognize it as an acid-base neutralization reaction, which produces a salt and water. Step 1: Identify reactants and products. Reactants: Sodium hydroxide (NaOH) and hydrochloric acid (HCl). Products: Sodium chloride (NaCl) and water (H_2O). Step 2: Write the unbalanced equation with state symbols. NaOH(aq) + HCl(aq) NaCl(aq) + H_2O(l) Step 3: Balance the equation. The equation is already balanced, with one sodium, one oxygen, two hydrogens, and one chlorine atom on both sides. The balanced chemical equation is: NaOH(aq) + HCl(aq) NaCl(aq) + H_2O(l) b) To calculate the molarity of the dilute HCl acid, we use the titration data and the stoichiometry of the reaction. Step 1: Identify the given values from the problem and the table. Molarity of NaOH (M_b) = 0.1 M Volume of NaOH (V_b) = 25 cm^3 (pipetted into the conical flask) From the table, the accurate titre readings are 27.1 cm^3 and 27.4 cm^3. Mean titre volume of HCl (V_a) = 27.1 cm^3 + 27.4 cm^32 = 54.5 cm^32 = 27.25 cm^3. From the balanced equation, the mole ratio of NaOH to HCl is 1:1. So, n_b = 1 and n_a = 1. Step 2: Use the titration formula. The formula for titration calculations is: (M_a V_a)/(n_a) = (M_b V_b)/(n_b) Where M_a and V_a are the molarity and volume of the acid, M_b and V_b are the molarity and volume of the base, and n_a and n_b are their respective stoichiometric coefficients from the balanced equation. Step 3: Substitute the values and solve for M_a. M_a = (M_b V_b n_a)/(V_a n_b) M_a = (0.1 M) × (25 cm^3) × 1(27.25 cm^3) × 1 M_a = (2.5)/(27.25) M M_a ≈ 0.091743 M Rounding to three significant figures: M_a = 0.0917 M c) To calculate the concentration of the dilute HCl acid in gdm^-3, we multiply its molarity by its molar mass. Step 1: Determine the molar mass of HCl. Atomic mass of H = 1.008 g/mol Atomic mass of Cl = 35.45 g/mol Molar mass of HCl = 1.008 + 35.45 = 36.458 g/mol Step 2: Calculate the concentration in gdm^-3. Concentration (gdm^-3) = Molarity (mol/dm^3) × Molar mass (g/mol) Using the unrounded molarity: Concentration = 0.091743 mol/dm^3 × 36.458 g/mol Concentration ≈ 3.344 g/dm^3 Rounding to three significant figures: Concentration = 3.34 gdm^-3 What's next?