Question

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 \(42.58 \mathrm{~mL}\) of \(0.2997 \mathrm{M}\) silver nitrate solution to reach the equivalence point in the titration. What is the mass percentage of chloride ion in the seawater if its density is \(1.025 \mathrm{~g} / \mathrm{mL} ?\)

Short answer

The mass percentage of chloride ions in the seawater sample is approximately \(1.765 \% \).

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation of the reaction between silver nitrate (AgNO₃) and chloride ions (Cl⁻) is as follows: \[AgNO_3(aq)+Cl^-(aq) \rightarrow AgCl(s)+NO_3^-(aq)\] Here, 1 mole of AgNO₃ reacts with 1 mole of Cl⁻ to form 1 mole of AgCl as the precipitate.

Calculate the moles of silver nitrate used in titration

We are given the volume and concentration of the silver nitrate solution, so we can calculate the moles of AgNO₃ used as follows: Moles of AgNO₃ = Volume × Concentration \[\text{Moles of AgNO₃} = 42.58 \mathrm{~mL} × 0.2997 \frac{\mathrm{mol}}{\mathrm{L}} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}} = 0.012765 \mathrm{~mol}\]

Calculate the moles of chloride ions

From the balanced chemical equation, we know that 1 mole of AgNO₃ reacts with 1 mole of Cl⁻. So, the moles of Cl⁻ in the seawater sample would be equal to the moles of AgNO₃ used in the titration. Moles of Cl⁻ = Moles of AgNO₃ \[\text{Moles of Cl⁻} = 0.012765 \mathrm{~mol}\]

Calculate the mass of chloride ions

Now that we have the moles of Cl⁻, we can find its mass using the molecular weight of chloride ions (MW of Cl = 35.45 g/mol). Mass of Cl⁻ = Moles of Cl⁻ × Molecular weight \[\text{Mass of Cl⁻} = 0.012765 \mathrm{~mol} × 35.45 \frac{\mathrm{g}}{\mathrm{mol}} = 0.4524 \mathrm{~g}\]

Calculate the mass of the seawater sample

We are given the volume and density of the seawater sample. We can calculate the mass of the sample as follows: Mass of seawater = Volume × Density \[\text{Mass of seawater} = 25.00 \mathrm{~mL} × 1.025 \frac{\mathrm{g}}{\mathrm{mL}} = 25.625 \mathrm{~g}\]

Calculate the mass percentage of chloride ions

Now we can calculate the mass percentage of chloride ions in the seawater sample using the mass of chloride ions and the mass of seawater: Mass percentage of Cl⁻ = (Mass of Cl⁻ / Mass of seawater) × 100 \[\text{Mass percentage of Cl⁻} = \frac{0.4524 \mathrm{~g}}{25.625 \mathrm{~g}} \times 100 = 1.765 \% \] So, the mass percentage of chloride ions in the seawater sample is approximately \(1.765 \% \).