Question

4.If 15.00 mL of 3.00 M potassium iodide is needed to reach the equivalence point with 10.00 mL of lead (Il) nitrate, determine the molarity of the lead (Il) nitrate solution

Answer 1

Answer:

2.25 M

Explanation:

The reaction that takes place is:

• 2KI + Pb(NO₂)₃ → PbI₂ + 2KNO₃

First we calculate how many potassium iodide moles reacted, using the given volume and concentration:

• 15.00 mL * 3.00 M = 45 mmol KI

Then we convert 45 mmoles of KI into mmoles of Pb(NO₂)₃, using the stoichiometric coefficients of the balanced reaction:

• 45 mmol KI * $\frac{1mmolPb(NO_3)_2}{2mmolKI}$ = 22.5 mmol Pb(NO₂)₃

Finally we calculate the molarity of the Pb(NO₂)₃ solution, using the calculated number of moles and given volume:

• 22.5 mmol Pb(NO₂)₃ / 10.00 mL = 2.25 M