Question

A student mixes 31.0 mL of 3.24 M Pb(NO3)2(aq) with 20.0 mL of 0.00217 M NaI(aq). How many moles of PbI2(s) precipitate from the resulting solution? What are the values of [Pb2 ], [I–], [NO3–], and [Na ] after the solution has reached equilibrium at 25 °C?

Answer 1

Given that ; 31.0 mL of 3.24 M Pb(NO3)2(aq) with 20.0 mL of 0.00217 M NaI(aq).

31.0 ml = 0.031 L

20.0 ml = 0.020 L

Total or final volume = 0.051 L

Number of mole s= molarity * volume in L

0.0310 liters x 3.24 moles Pb+2/1 liter = 0.10044 moles Pb+2
0.0200 liters x 0.00217 moles I-/1 liter = 4.34 x 10^-5 moles I-

Pb+2 + 2I- ===⇒ PbI2
Limiting reagent is the I- ,

I- is limiting agent due to following reasons:

1. It completely reacted in the reaction.
2. It determines the amount of the product in mole.

4.34 x 10^-5 moles I- x 1mole PbI2/2 moles I- = 2.17 x 10^-5 moles of PbI2 precipitated

Final volume of solution = 0.051 liters

Moles of Pb+2 remaining in solution = 0.10044 – 4.34 x 10^-5 ; essentially 0.1003966 moles of Pb+2 remain
[Pb+2] = 0.1003966 /0.051 liters = 1.97

Ksp = [Pb+2][I-]^2 = 9.8 x 10^-9
[1.97][I-]^2= 9.8 x 10^-9
[I-]^2 = 9.8 x 10^-9/1.97
[I-] = 7.05 x 10^-5

moles NO3^-1 = 0.031 liters x 3.24 moles Pb(NO3)2/1 liter x 2 moles NO3^-1/1 mole Pb(NO3)2 = 0.20088 moles NO3^-1

[NO3^-1] = 0.20088 moles/0.051 liters = 3.94 Molar

0.00217 moles/liter x 1 mole Na+/1mole NaI x 0.020 liters = 0.0000434 moles I-

[Na+] = 0.0000434 moles Na+/0.051 liters = 0.000851