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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Question 4(a): Number of moles of HCl used Step 1: Formula for number of moles n = c × V where n is number of moles (mol), c is concentration (mol dm^-3), V is volume (dm^3). Step 2: Convert volume to dm^3 V = (25)/(1000) = 0.025\ dm^3 Step 3: Substitute values n(HCl) = 0.1\ mol dm^-3 × 0.025\ dm^3 = 0.0025\ mol 0.0025 mol Question 4(b): Number of moles of NaOH that reacted with HCl Step 1: Moles of HCl used in back titration (equals moles of excess NaOH, since 1:1 reaction) n(HCl, back) = 0.1\ mol dm^-3 × (24.9)/(1000)\ dm^3 n(HCl, back) = 0.1 × 0.0249 = 0.00249\ mol Excess NaOH = 0.00249\ mol. Step 2: Total moles of NaOH added n(NaOH, total) = n(HCl, pipetted) + n(HCl, back) n(NaOH, total) = 0.0025\ mol + 0.00249\ mol = 0.00499\ mol Step 3: Moles of NaOH that reacted with HCl n(NaOH, reacted) = n(NaOH, total) - excess NaOH n(NaOH, reacted) = 0.00499\ mol - 0.00249\ mol = 0.0025\ mol 0.0025 mol Question 4(c): Concentration of NaOH in moles per dm^3 Step 1: Volume of NaOH added in dm^3 V(NaOH) = (25.0)/(1000) = 0.0250\ dm^3 Step 2: Concentration formula c = (n)/(V) Step 3: Substitute values c(NaOH) = 0.00499\ mol0.0250\ dm^3 = 0.20\ mol dm^-3 0.20 mol dm^-3