The Ksp for PbI2 is 9.8×10−9 . Will PbI2 precipitate if 2.5 mL of 0.0025 M...

Question

Question

The Ksp for PbI2 is 9.8×10−9 . Will PbI2 precipitate if 2.5 mL of 0.0025 M Pb(NO3)2 is added to 7.5 mL of 0.0050 M NaI?

Answer 1

Answer #1

total volume of solution mixture = (2.5 + 7.5) = 10.0 ml

2.5 mL of 0.0025 M Pb(NO3)2 is added to 7.5 mL of 0.0050 M NaI mixed.

[Pb^2+] = [Pb(NO3)2] = 2.5 mL * 0.0025 M / 10.0 ml = 6.25 * 10^-4 M

[I-] = [NaI] = 7.5 mL * 0.0050 M / 10.0 ml = 3.75 * 10^-3 M

PbI2 ............> Pb^2+ + 2 I-

Q = [Pb^2+][I-]^2 = 6.25 * 10^-4 * (3.75 * 10^-3)^2 = 8.79 * 10^-9

we have

Ksp for PbI2 is 9.8×10−9

thus

Q < Ksp

so

PbI2 do not precipitated.

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Answer 2

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