20.0 mL of a 2.2 M lead (II) nitrate solution is mixed with 100.0 mL of a 0.75 M magnesium sulfate solution.
What is the concentration of nitrate ions in the resulting solution?
First, calculate the number of moles of NO₃⁻.
n = (0.0200 L)(2.2 M Pb(NO₃)₂) = 0.044 mol Pb(NO₃)₂
Since there are two moles of NO₃⁻ per one mole of Pb(NO₃)₂ that dissociates, the umber of moles of NO₃⁻ is double the number of moles of Pb(NO₃)₂. Thus...
n = 0.088 mol NO₃⁻
And, assuming the volumes are additive...
M = (0.088 mol) / (0.02 + 0.100) = 0.73 M NO₃⁻
To find the concentration of nitrate ions in the resulting solution, we need to determine the number of moles of nitrate ions in both the lead (II) nitrate and magnesium sulfate solutions, and then calculate the concentration of nitrate ions in the final solution.
First, let's calculate the number of moles of nitrate ions in the lead (II) nitrate solution:
Molarity is defined as moles of solute per liter of solution. We have 20.0 mL of a 2.2 M lead (II) nitrate solution, which can be converted to liters by dividing by 1000:
20.0 mL / 1000 = 0.020 L
Now, we can calculate the number of moles of nitrate ions in the lead (II) nitrate solution using the formula:
moles = concentration (M) x volume (L)
moles = 2.2 M x 0.020 L = 0.044 mol
Next, let's calculate the number of moles of nitrate ions in the magnesium sulfate solution:
Similarly, we have 100.0 mL of a 0.75 M magnesium sulfate solution, which can also be converted to liters:
100.0 mL / 1000 = 0.100 L
Using the same formula, we can calculate the number of moles of nitrate ions in the magnesium sulfate solution:
moles = 0.75 M x 0.100 L = 0.075 mol
Now, let's calculate the total number of moles of nitrate ions in the final solution by adding the moles from both solutions:
0.044 mol + 0.075 mol = 0.119 mol
Finally, let's calculate the concentration of nitrate ions in the resulting solution:
Concentration is defined as moles of solute per liter of solution. In this case, we have a total of 0.119 moles of nitrate ions in a final solution volume of 100.0 mL + 20.0 mL = 120.0 mL, which can be converted to liters by dividing by 1000:
120.0 mL / 1000 = 0.120 L
Therefore, the concentration of nitrate ions in the resulting solution is:
concentration = moles / volume = 0.119 mol / 0.120 L = 0.992 M
Therefore, the concentration of nitrate ions in the resulting solution is 0.992 M.