Question

How many milliliters of a 0.160 M potassium chloride solution should be added to 49.0 mL of a 0.270 M lead(II) nitrate solution to precipitate all of the lead(II) ion?

Answer 1

Approx....`170*mL`....

Explanation:

`Pb^(2+) + 2Cl^(-) rarr PbCl_2(s)darr`

`"Moles of lead"=0.270*mol*L^-1xx49.0xx10^-3*L=0.01323*mol..`

And we require TWO EQUIV of chloride ion....

`=(2xx0.01323*mol)/(0.160*mol*L^-1)xx10^3*mL*L^-1=165.4*mL`