Question

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?

Answer 1

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⁻