The mass percentage of chloride ion in a 25.00-mL sample of seawater was determined by titrating the sample with silver nitrate, precipitating silver chloride. It took 42.58 ml of 0.2997 M silver nitrate solution to reach the equivalence point in the titration. What is the mass percentage of chloride ion in seawater if its density is 1.025 g/ml?
Solution: Here, we are going to calculate the mass percentage of chloride ion in water with the help of data given. Step1: Given, volume of silver nitrate solution used = 42.58 mL = 10008L = 0.04258 L Molarity of silver nitrate solution = 0.2997 M Therefore, number of moles of silver nitrate = volume of solution in litres x molarity = 0.04258 L x 0.2997 M = 0.013 mol Step2: Now, the reaction between silver nitrate and chloride ion takes place in the following manner: - - Cl(aq) + AgNO (aq3-------> AgCl(s) + NO (aq) 3 From the above reaction, it is clear that 1 mol of AgNO reacts with 3 mol of chloride ion for complete precipitation. Therefore, 0.013 mol of AgNO will react 3ith 0.013 mol of chloride ion for complete precipitation. Step3: Now, 1 mol of chloride ion = molar mass of Cl - - - Therefore, 0.013 mol of Cl = 0.013 x molar mass of Cl = 0.013 x 35.45 g [molar mass of Cl = 35.45 g/mol] = 0.46 g Step4: Again, density can be defined as mass per unit volume. i.e., Density = mass / volume Therefore, mass = density x volume Given, density of the seawater sample = 1.025 g/mL Volume of the seawater sample = 25.00 mL Therefore, mass of the seawater sample = 1.025 x 25.00 [(g/ml) x mL] = 25.625 g Step5: Mass percentage of Chloride ion in the sample is given by: Mass % of Cl = (mass of Cl / total mass of sample) x 100% = (0.46 / 25.625) x 100% = 0.0179 x 100% =1.79 % Thus, mass percentage of chloride ion in the seawater sample is 1.79 %.