Calculate the volume of 2,180 ppm Pb(NO3)2 solution required to prepare 1.2550 L of a 0.0021 M Pb2+ solution.
2,180 ppm = 2,180 mg Pb(NO3)2/L soln = 2.18 g Pb(NO3)2/L soln
moles Pb(NO3)2 = 2.18/molar mass Pb(NO3)2 = about 0.0066 M.
Then use c1v1 = c2v2 to see how much to dilute this solution.
This calculation is based on the density of the 2180 ppm soln being 1.00 g/mL.
how would i use 2180 as a c1 or c2?
To use my approximate values.
c2 = 0.0021 M soln you want
v2 = 1.2550 L soln you want
c1 = 0.0066 M soln you have
V1 = ? = want to know this
To calculate the volume of the 2,180 ppm Pb(NO3)2 solution required to prepare 1.2550 L of a 0.0021 M Pb2+ solution, we need to use some conversion factors and molar ratios.
First, let's understand the relationship between ppm (parts per million) and molarity.
Parts per million (ppm) is a unit used to describe the concentration of a solute in a solution. It represents the number of parts of solute divided by the total number of parts in the solution, multiplied by 10^6. Mathematically, we can express ppm as follows:
ppm = (mass of solute / mass of solution) x 10^6
Now, we can use the given information to calculate the volume of the Pb(NO3)2 solution required.
Step 1: Calculate the mass of Pb(NO3)2 in the desired concentration:
Since the desired concentration is 0.0021 M, we can use the molar mass of Pb(NO3)2 to calculate the mass of Pb(NO3)2 in 1.2550 L of solution.
The molar mass of Pb(NO3)2 = molar mass of Pb + 2(molar mass of NO3)
= 207.2 g/mol + 2(3 x 16.0 g/mol)
= 331.2 g/mol
Mass of Pb(NO3)2 = molarity x volume x molar mass
= 0.0021 M x 1.2550 L x 331.2 g/mol
= 0.549 g
Step 2: Calculate the volume of 2,180 ppm Pb(NO3)2 solution:
Now, using the mass of Pb(NO3)2 calculated above, and the ppm value of 2,180 ppm, we can determine the volume of the Pb(NO3)2 solution.
ppm = (mass of solute / mass of solution) x 10^6
2,180 ppm = (0.549 g / volume of solution) x 10^6
Solving the equation for the volume of the solution:
volume of solution = (0.549 g / 2,180 ppm) x 10^6
= 0.252 L
So, the volume of the 2,180 ppm Pb(NO3)2 solution required to prepare 1.2550 L of a 0.0021 M Pb2+ solution is approximately 0.252 L.