A solution contains 0.0150 M lead(II) ion. A concentrated sodium iodide solution

Question

A solution contains 0.0150 M lead(II) ion. A concentrated sodium iodide solution is added dropwise to precipitate lead iodide (assume no volume change). K_sp for PbI_2 is 6.5 times 10^-9. (Nal is a 100% soluble salt. This is a precipitation problem.) 2Nal(aq) + Pb^2+(aq) rightarrow Pbl_2(s) + 2Na^+(aq) Pbl_2 rightarrow Pb^2+(aq) + 2F^-(aq) At what concentration of I^- does precipitate start to form? When [I^-] = 2.0 times 10^-3, what is the lead-ion concentration? What percentage of the lead(II) originally present remains in solution?

Explanation / Answer

a. Ksp for PbI2 = [Pb2+][I-]2

with,

[Pb2+] = 0.0150 M

we get [I-] concentration required to start precipitation,

[I-] = sq.rt.(Ksp/[Pb2+]

     = sq.rt.(6.5 x 10^-9/0.0150)

     = 6.583 x 10^-4 M

b. When [I-] = 2.0 x 10^-3 M

[Pb2+] = Ksp/[I-]^2

            = 6.5 x 10^-9/( x 10^-3)^2

            = 1.625 x 10^-3 M

% of Pb2+ remaining in solution = (0.0150 - 1.625 x 10^-3)/0.0150) x 100 = 89.167%

  

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