Answer to Question #155412 in General Chemistry for C

Question

Can we expect a precipitate of lead (II) iodide, PbI2, to form when 20.00 mL of 0.0050 mol/L 

     aqueous calcium iodide, CaI2, solution is added to 80.0 mL of 0.0010 mol/L aqueous solution of  lead(II) nitrate, Pb(NO3)2?  The Ksp for lead (II) iodide is 1.4X10-8. 

Answer 1

The ionic equation for Cal, is as follows:

$CaI_{(aq)}→ 2I^{-}_{(aq)} + Ca^{2+}_{(aq)}$

The ionic equation for Pb(NO₃)₂, is as follows:

Pb(NO₃)_2(aq) → Pb²⁺_(aq) + 2NO^-_{3 (aq)}

The ionic equation for solubility is as follows:

Pb(NO₃)_2(aq) + Cal_(aq) → PbI_(s)+ Ca(NO₃)_2(aq)

The reaction quotient Q is:

Q = [Pb2+][I-]²

The initial concentration of Pb²⁺ is as follows:

Initial concentration of [Pb²⁺] = $\dfrac{Molarity×Volume}{Total \ volume}$ = $\dfrac{0.0010M × 80.0 mL}{80.0mL+20.0 mL}$ = 1.0×10^-3M

The reaction quotient Q will be as follows:

Q = [Pb²⁺][I-]²

= [8x10^-4] 1.0x10^-3]²

= 8.0 x10^-10M

The solubility product of Pbl₂ is 1.4 x10^-8 and the reaction quotient is 8.0x10^-10 is less than the Ksp

So, K_sp > Q therefore precipitation does not occur