Chemistry Homework

Question

How many milliliters of a 0.180 M potassium chloride solution should be added to 24.0 mL of a 0.330 M lead(II) nitrate solution to precipitate all of the lead(II) ion?

Answer 1

First write the balanced equation:

2 KCl + Pb(NO₃)₂ -> 2 KNO₃ + PbCl₂

Then convert mL of lead(II) nitrate to moles of lead(II) nitrate:

24.0mL * (0.330mole / 1000mL) = 0.00792 moles of lead(II) nitrate

From the balanced equation, one mole of lead(II) nitrate reacts with two moles of potassium chloride:

0.00792 moles of lead(II) nitrate * (2 moles of potassium chloride / 1 mole of lead(II) nitrate) = 0.0158 moles of potassium chloride

Finally, convert moles of potassium chloride to mL of potassium chloride:

0.0158 moles of potassium chloride * (1000mL / 0.180 mole) = 87.8 mL of potassium chloride (88.0 mL if you did not round in the second step)