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Chapter 4: Problem 143

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

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

Step by step solution

01

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
02

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
03

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
04

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.

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