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 calculations for the standardization of sodium hydroxide using sulphamic acid. a) To find the concentration of the Sulphamic acid solution and the concentration of the NaOH solution in moldm^-3. Step 1: Calculate the concentration of the Sulphamic acid solution. From Table 1, the mass of Sulphamic acid used is 2.36 g. The relative molecular mass of Sulphamic acid (NH_2SO_3H) is given as 97.1 g/mol. The procedure states that the acid was dissolved and "made up to the mark", implying a standard solution was prepared. A common volume for such preparations is 250 cm^3. Volume of Sulphamic acid solution = 250 cm^3 = 0.250 dm^3. Moles of Sulphamic acid = MassMolar mass Moles of Sulphamic acid = 2.36 g97.1 g/mol ≈ 0.0243048 mol Concentration of Sulphamic acid (M_a) = MolesVolume (dm^3) M_a = 0.0243048 mol0.250 dm^3 ≈ 0.0972192 mol/dm^3 Rounding to three significant figures: Concentration of Sulphamic acid = 0.0972 mol/dm^3 Step 2: Calculate the concentration of the NaOH solution in moldm^-3. The balanced chemical equation for the reaction is: NaOH(aq) + NH_2SO_3H(aq) NH_2SO_3Na(aq) + H_2O(l) Sulphamic acid is a monobasic acid, so the mole ratio of NaOH to NH_2SO_3H is 1:1. Thus, n_b = 1 (for NaOH) and n_a = 1 (for Sulphamic acid). From the titration data (Table 2): Volume of Sulphamic acid (V_a) = 25 cm^3 (pipetted into the conical flask) Mean titre volume of NaOH (V_b) = 25.2 cm^3 (average of the two accurate readings) Using the titration formula: (M_a V_a)/(n_a) = (M_b V_b)/(n_b) Rearranging to solve for M_b: M_b = (M_a V_a n_b)/(V_b n_a) Substitute the values (using the unrounded M_a for accuracy): M_b = (0.0972192 mol/dm^3) × (25 cm^3) × 1(25.2 cm^3) × 1 M_b = (2.43048)/(25.2) mol/dm^3 M_b ≈ 0.0964476 mol/dm^3 Rounding to three significant figures: Concentration of NaOH = 0.0964 mol/dm^3 b) To calculate the concentration of the NaOH solution in gdm^-3. Step 1: Determine the molar mass of NaOH. Atomic mass of Na = 22.99 g/mol Atomic mass of O = 16.00 g/mol Atomic mass of H = 1.008 g/mol Molar mass of NaOH = 22.99 + 16.00 + 1.008 = 39.998 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 of NaOH: Concentration = 0.0964476 mol/dm^3 × 39.998 g/mol Concentration ≈ 3.8577 g/dm^3 Rounding to three significant figures: Concentration of NaOH = 3.86 gdm^-