Question

You made 100.0 mL of a lead(II) nitrate solution for lab but forgot to cap it. The next lab session you noticed that there was only 80.0 mL left (the rest had evaporated). In addition, you forgot the initial concentration of the solution. You decide to take 2.00 mL of the solution and add an excess of a concentrated sodium chloride solution. You obtain a solid with a mass of 3.407 g. What was the concentration of the original lead(II) nitrate solution?

Short answer

The concentration of the original lead(II) nitrate solution was approximately \(6.125 \, mol/L\).

Step-by-step solution

Determine the moles of lead(II) chloride

Since we have the mass of lead(II) chloride (PbCl2) precipitate (3.407 g), we can calculate the moles of PbCl2 using its molar mass (Pb: 207.2 g/mol, Cl: 35.45 g/mol): Moles of PbCl2 = 3.407 g / (207.2 g/mol + 2 * 35.45 g/mol) = 3.407 g / 278.1 g/mol ≈ 0.01225 mol

Calculate the moles of lead(II) nitrate in the 2 mL solution

Since the reaction between lead(II) nitrate and sodium chloride is a 1:1 stoichiometric reaction, the moles of lead(II) nitrate (Pb(NO3)2) in the 2 mL solution will be equal to the moles of lead(II) chloride (PbCl2) obtained: Moles of Pb(NO3)2 in 2 mL solution = 0.01225 mol

Adjust the moles of lead(II) nitrate to the original 100 mL solution

Since the total volume of lead(II) nitrate solution was 100 mL at the beginning, and 20 mL of the solution had evaporated, leaving 80 mL, we need to find the moles of Pb(NO3)2 in the original 100 mL solution: Moles of Pb(NO3)2 in original 100 mL solution = (0.01225 mol) * (100 mL / 2 mL) = 0.6125 mol

Calculate the concentration of the original lead(II) nitrate solution

Now we can calculate the concentration of the original lead(II) nitrate solution, considering that the total volume of solution was 100 mL or 0.100 L: Concentration of Pb(NO3)2 = Moles of Pb(NO3)2 / Volume of solution (L) Concentration of Pb(NO3)2 = 0.6125 mol / 0.100 L ≈ 6.125 mol/L The concentration of the original lead(II) nitrate solution was approximately 6.125 mol/L.

Key concepts

Molar Mass Calculation

To determine the moles of a compound formed in a chemical reaction, it's crucial to know its molar mass. Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It involves summing up the atomic masses of all the atoms in the molecule.

For example, to find the molar mass of lead(II) chloride (PbCl₂), we add:

• Lead (Pb): 207.2 g/mol
• Chlorine (Cl): 35.45 g/mol (and since there are two chlorine atoms, multiply by 2)

This results in a molar mass for PbCl₂ of 278.1 g/mol. Using this molar mass, you can convert the mass of a substance to moles. For instance, if you have 3.407 g of PbCl₂, divide this mass by 278.1 g/mol to find the number of moles:\[\text{Moles of PbCl₂} = \frac{3.407 \text{ g}}{278.1 \text{ g/mol}} \approx 0.01225 \text{ mol}\]This calculation is foundational for assessing how substances interact in a reaction.

Concentration Determination

Concentration tells us how much of a solute is present in a solution. It is typically expressed in moles per liter (mol/L), also known as molarity. To calculate the concentration, divide the number of moles of the solute by the total volume of the solution in liters.

In the exercise, we determined the moles of lead(II) nitrate in a 2 mL sample from the total solution. Since we found that the moles of lead(II) chloride obtained was 0.01225 mol, which equals the moles of lead(II) nitrate due to the 1:1 stoichiometry, we adjusted this value to correspond with the original 100 mL solution:\[\text{Moles of Pb(NO₃)₂ in 100 mL} = 0.01225 \text{ mol} \times \frac{100 \text{ mL}}{2 \text{ mL}} = 0.6125 \text{ mol}\]Using this information, we find the concentration of the original solution:\[\text{Concentration} = \frac{0.6125 \text{ mol}}{0.100 \text{ L}} \approx 6.125 \text{ mol/L}\]This value reflects how much lead(II) nitrate was present in each liter of the original solution.

Chemical Reaction Stoichiometry

Stoichiometry is the method of using balanced chemical equations to determine the relationships between reactants and products. It is essential in predicting the outcomes and quantities needed in reactions.

In chemical equations, coefficients indicate the ratio of moles of each substance involved in the reaction. For instance, in the reaction between lead(II) nitrate (Pb(NO₃)₂) and sodium chloride (NaCl) to form lead(II) chloride (PbCl₂), the reaction is expressed as:\[\text{Pb(NO₃)₂ (aq) + 2 NaCl (aq) → PbCl₂ (s) + 2 NaNO₃ (aq)}\]This equation shows a 1:1 molar ratio between Pb(NO₃)₂ and PbCl₂. Thus, the moles of PbCl₂ formed directly indicate the moles of Pb(NO₃)₂ initially present.

Understanding stoichiometry aids in making precise calculations in laboratory settings, helping us to predict the needed quantities and resulting products of our reactions.