Question

To verify the previous experiment, the chemistry student carried out the next experiment and recorded the data for you in the following Data Table.

Directions: Place 100 mL of filtered, saturated PbCl, in a 150 ml beaker and put it under a dryer to evaporate the water. Record the necessary data in the Data Table.

Complete the Data Table.

Since the data allows it, you should use 3 significant digits for all your answers.

Molecular weight of PbC12-278 g/mole

Mass of beaker + PbCl₂-62.19 g

Mass of beaker-57.77 €

Mass of PbCl₂ per 100 mL-Choose...

Moles PbCl₂ per 100 mL-Choose...

[Pb ] based upon this investigation - Choose M

[CI] based upon this investigation - Choose... M

This is close to our answer of 0.34 M above. The difference may be due to experimental error and to rounding numbers off because we were limited to 2 significant digits in the first
part of this experiment. This illustrates the importance of using care with our procedures in the laboratory and to record instrument readings to as many significant digits as
possible.

Answer 1

based on the calculations I provided using the given values for the molecular weight of PbCl₂, the mass of the beaker and PbCl₂, and the mass of the beaker, the concentration of Pb²⁺ ions is 0.159 M and the concentration of Cl⁻ ions is 0.318 M. These values differ slightly from the values given earlier due to the use of additional significant digits in the calculations.

Step-by-step explanation:

To find the concentration of Pb²⁺ and Cl⁻ ions, we need to use the given information about the molecular weight of PbCl₂ and the mass of PbCl₂ per 100 mL of solution.

First, we use the mass of the beaker and PbCl₂, and the mass of the beaker, to calculate the mass of PbCl₂:

Mass of PbCl₂ = (mass of beaker + PbCl₂) - (mass of beaker)

Mass of PbCl₂ = 62.19 g - 57.77 g = 4.42 g

Next, we need to convert this mass to grams per 100 mL:

Mass of PbCl₂ per 100 mL = (mass of PbCl₂ / volume of solution) x (100 mL / 1 L)

Mass of PbCl₂ per 100 mL = (4.42 g / 100 mL) x (100 mL / 1 L) = 44.2 g/L

We also need to convert this mass to moles per 100 mL. To do this, we need to use the molecular weight of PbCl₂:

Moles of PbCl₂ per 100 mL = (mass of PbCl₂ / molecular weight of PbCl₂) x (100 mL / 1 L)

Moles of PbCl₂ per 100 mL = (4.42 g / 278 g/mol) x (100 mL / 1 L) = 0.0159 mol/L

This gives us the moles of PbCl₂ per 100 mL of solution. To find the concentration of Pb²⁺ ions, we use the stoichiometry of the reaction between PbCl₂ and water:

PbCl₂ + 2H₂O → Pb²⁺ + 2Cl⁻ + 2H⁺

For every 1 mole of PbCl₂ that dissolves in water, we get 1 mole of Pb²⁺ ions. Therefore, the concentration of Pb²⁺ ions is the same as the concentration of PbCl₂ in the solution:

[Pb²⁺] = moles of PbCl₂ per 100 mL = 0.0159 mol/L = 0.159 M

Finally, to find the concentration of Cl⁻ ions, we use the stoichiometry of the reaction again:

PbCl₂ + 2H₂O → Pb²⁺ + 2Cl⁻ + 2H⁺

For every 1 mole of PbCl₂ that dissolves in water, we get 2 moles of Cl⁻ ions. Therefore, the concentration of Cl⁻ ions is twice the concentration of PbCl₂ in the solution:

[Cl⁻] = 2 x moles of PbCl₂ per 100 mL = 2 x 0.0159 mol/L = 0.0318 M

Note that the concentrations of Pb²⁺ and Cl⁻ ions are slightly different from the values given earlier due to the use of additional significant digits in the calculations.