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In order to calculate the $K_\mathrm{sp}$ of lead(II) iodide, I mixed $\pu{6.0 mL}$ of $\pu{0.01 M}$ $\ce{Pb(NO3)2}$, $\pu{6.0 mL}$ of $\pu{0.02 M}$ $\ce{KI}$, and $\pu{8.0 mL}$ of water, and a precipitate (barely) formed.

From this, I calculated $[\ce{Pb^2+}]$ to be $\pu{0.0036 M}$ and $[\ce{I-}]$ to be $\pu{0.0060 M}$, giving

$$K_\mathrm{sp} = [\ce{Pb^2+}][\ce{I-}]^2 = 1.1 \times 10^{-7}.$$

However, the actual value (at $\pu{25 ^\circ C}$) is $7.9\times 10^{-9}$ (Skoog et al. Fundamentals of Analytical Chemistry), so my result is off by a factor of $\sim 10$.

I'm pretty sure I had clean equipment and measured carefully and accurately. Why is there a discrepancy? Is my maths wrong, or are there some inaccuracies in my experiment?

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    $\begingroup$ I don't understand your issue… “barely” is not helpful: your product of concentration is higher than the solubility product, so it precipitates, which is what you expected. I suppose “barely” means you expected a larger mass of precipitate, but unless you actually weighted it and found it to be inferior to the expected calculated mass, I don't see the issue… $\endgroup$ – F'x May 6 '12 at 20:02
  • $\begingroup$ @F'x I actually did a variety of concentrations to narrow it down to the point at which precipitation could be observed, varying my solutions by 1 mL. So 'barely' here means 'as close to equilibrium as I could repeat with the solutions on hand'. $\endgroup$ – Henry May 8 '12 at 18:24
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It's probably an experimental error. The 'barely' may be important here.

For one, you may not be sure how accurate your initial molarities are. A mistake in one of them ruins it all. Titrate them properly and check. Titration usually gives two significant digits.

Also, since its a cube overall, a small mistake can get much larger. You may want more accuracy in your readings (one significant digit isn't good enough), and you need to improve on the 'barely'. Though a factor of 11 needs a variation by a factor of 2.3 in both readings. So I don't know.

Have you redone the experiment? Doing an experiment more than once is integral to getting good values.

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I don't have time to do the calculations manually, but inputing your problem data into a chemical equilibrium software (here Dozzaqueux) reveals that the precipitate you observe is Pb(OH)2, while no PbI2 should form. The former is white, while the later is yellow; can you tell us what color was your precipitate?

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  • $\begingroup$ It was white, and now I'm definitely confused! :) Thanks though, I will likely learn from this too $\endgroup$ – Henry May 8 '12 at 18:24
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How long you observed your mixture? It is possible that you had not reached the equilibrium required for the solubility product constant to be valid. Allowing time for the system to reach equilibrium might have resulted in the appearance of more precipitate that would indicate that you had added too much material.

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