Question

Consider these mixtures:
I. 100. mL of 0.06 M Pb(NO₃)₂ and 50.0 mL of 0.03 M NaBr
II. 100. mL of 0.008 M Pb(NO₃)₂ and 100.0 mL of 0.05 M NaBr
The K_sp of PbBr₂ = 6.6 x 10^-6.

How do I found out if a precipitate will form for each?

Answer 1

If

ionic product > Ksp

Then precipitate will form

i) I. 100. mL of 0.06 M Pb(NO₃)₂ and 50.0 mL of 0.03 M NaBr

[Pb⁺²] = intitial concentration X initial volume /total volume = 0.06 X 100 / 150 = 0.04 M

[Br-] = intitial concentration X initial volume /total volume = 0.03 X 50 / 150 = 0.01 M

I. P = [Pb⁺²][Br-]² = 0.04 X (0.01)² = 4 X 10^-6

K_sp of PbBr₂ = 6.6 x 10^-6 > I.P ,so no ppt will form

ii) 100. mL of 0.008 M Pb(NO₃)₂ and 100.0 mL of 0.05 M NaBr

[Pb⁺²] = intitial concentration X initial volume /total volume = 0.008 X 100 / 250 = 0.004 M

[Br-] = intitial concentration X initial volume /total volume = 0.05 X 100 / 200 = 0.025 M

I. P = [Pb⁺²][Br-]² = 0.004 X (0.025)² = 2.5 X 10^-6

K_sp of PbBr₂ = 6.6 x 10^-6 > I.P ,so no ppt will form

NOTE: Please check the concentration of Pb(NO₃)₂ in question , it is 0.008 M so the solution is accordingly