The mass percentage of chloride ion in a 29.00 sample of seawater was determined by titrating the sample with silver nitrate, precipitating silver chloride. It took 42.42 of 0.2998 silver nitrate solution to reach the equivalence point in the titration.
29.00 WHAT?
What s the question?
To find the mass percentage of chloride ion in the seawater sample, we need to calculate the amount of chloride ion present in the sample and then express it as a percentage of the sample mass.
Let's break down the problem step by step:
1. Calculate the moles of silver nitrate used in the titration:
Moles of silver nitrate = volume of silver nitrate solution used (in L) × molarity of silver nitrate solution
From the given information:
Volume of silver nitrate solution used = 42.42 mL = 0.04242 L
Molarity of silver nitrate solution = 0.2998 M
Moles of silver nitrate = 0.04242 L × 0.2998 M = 0.012714 moles
2. Determine the moles of chloride ions present in the sample:
According to the balanced chemical equation for the reaction between silver nitrate and chloride ion, 1 mole of silver nitrate reacts with 1 mole of chloride ion to form 1 mole of silver chloride.
Moles of chloride ions = Moles of silver nitrate used in titration = 0.012714 moles
3. Calculate the mass of chloride ions in the sample:
To find the mass of chloride ions, we need to know the molar mass of chloride. The molar mass of chloride (Cl-) is 35.45 g/mol.
Mass of chloride ions = Moles of chloride ions × Molar mass of chloride
Mass of chloride ions = 0.012714 moles × 35.45 g/mol = 0.4516 g
4. Calculate the mass percentage of chloride ion in the sample:
Mass percentage of chloride ion in the sample = (Mass of chloride ions / Mass of sample) × 100%
From the given information:
Mass of sample = 29.00 g
Mass percentage of chloride ion = (0.4516 g / 29.00 g) × 100%
Mass percentage of chloride ion = 1.557%
Therefore, the mass percentage of chloride ion in the seawater sample is approximately 1.557%.