Question

Calcium hypochlorite (Ca(OCl)2, MW = 142.983 g/mol) is often used as the source of the hypochlorite ion (OCl–, MW = 51.452 g/mol) in solutions used for water treatment. A student must prepare 50.0 mL of a 55.0 ppm OCl– solution from solid Ca(OCl)2, which has a purity of 94.0%.

A) Calculate the mass of the impure reagent required to prepare the solution. Assume the density of the solution is 1.00 g/mL.

B) Which of the following methods would work best to accurately prepare 50.0 mL of the 55.0 ppm OCl– solution?

Use an analytical balance to weigh out an amount of reagent larger than what was determined in part A. In a volumetric flask, combine the reagent and water to create a solution with a concentration greater than 55.0 ppm. Use a pipet to transfer a portion of the concentrated solution to a separate 50.0 mL volumetric flask and dilute to the line with water to create a solution with a concentration of 55.0 ppm.

Use an analytical balance to weigh out an amount of reagent larger than what was determined in part A. In a volumetric flask, combine the reagent and water to create a solution with a concentration greater than 55.0 ppm. Use a graduated cylinder to transfer a portion of the more concentrated solution to a 50.0 mL volumetric flask and dilute to the line with water to create a solution with a concentration of 55.0 ppm.

Use an analytical balance to weigh out the amount of reagent determined in part A. Transfer the reagent to a 50.0 mL volumetric flask, add water to the line, and mix thoroughly.

Use an analytical balance to weigh out the amount of reagent determined in part A. Then use a graduated cylinder to measure out 50.0 mL of water. Combine the reagent and water in a 50.0 mL volumetric flask and mix it thoroughly.

Answer 1

A) 4.06 mg

B) Use an analytical balance to weigh out the amount of reagent determined in part A. Transfer the reagent to a 50.0 mL volumetric flask, add water to the line, and mix thoroughly.

Step-by-step explanation:

A) When the calcium hypochlorite is added to the water, it will dissolve, and will produce the hypochlorite ion by the reaction:

Ca(OCl)₂(s) → Ca⁺²(aq) + 2OCl⁻(aq)

Thus, 1 mol of Ca(OCl)₂(s) produces 2 moles of OCl⁻.

The concentration of the solution is 55.0 ppm, which means that there are 55.0g of OCl⁻ per 1,000,000 g of the solution, thus, if we multiply this by the density of the solution (1.00 g/mL), we will have the concentration by w/v (weight per volume) of the hypochlorite:

55.0 g of OCl⁻/1,000,000 *1 g/mL

5.5x10⁻⁵ g OCl⁻/mL

Thus, the mass of the OCl⁻ presented in 50.0 mL of the solution is 2.75x10⁻³ g (5.5x10⁻⁵ g OCl⁻/mL * 50 mL). The number of moles of the substance is the mass divided by its molar mass:

nOCl⁻ = 2.75x10⁻³/51.452

nOCl⁻ = 5.345x10⁻⁵ mol

So, if 2 moles of OCl⁻ is provided by 1 mol of Ca(OCl)₂, then, the number of moles needed is half of the number of moles of the ion:

nCa(OCl)₂ = 5.345x10⁻⁵/2

nCa(OCl)₂ = 2.67x10⁻⁵ mol

The mass necessary of it is the number of moles multiplied by its molar mass:

m = 2.67x10⁻⁵ * 142.983

m = 3.82x10⁻³ g

m = 3.82 mg

Because the reagent is impure, the mass needed will be higher, because only the Ca(OCl)₂ will react. The purity is the mas off the reagent divided by the total mass:

3.82/M = 0.94

0.94M = 3.82

M = 4.06 mg

B) To best prepare, the solution, the mass calculated of the reactant must be weighed out in an analytical balance, because in this balance the error value is low (it has 5 decimal figures). Then, the reagent must be put in a 50.0 mL flask, because it's the solution volume we want, and in the volumetric flask, the volume is trustworthy. Then, the water is added until the line, which represents the total volume (50.0 mL).

When this exact mass is added to this volume, the solution must be mixed until it becomes homogenous. When it happens, the reagent dissociates in the ion OCl⁻, and the steps taken in part A occur in inverse order, given the most accurate concentration.