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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Boxxxed on March 18, 2011, 08:30:17 PM

Title: Solubility, How much precipitates?
Post by: Boxxxed on March 18, 2011, 08:30:17 PM

If I have Q and ksp and Q greater than ksp can I tell how much precipitated?

Q=(0.000952)(0.0019) = 1.13x10^-6

ksp = 1.8x10^-10

The question just asks if it will precipitate, which is obvious but I am curious to know if this is enough information to find out exactly how much precipitated.

Sqroot Q = 0.00106
Sqrootksp= 0.000013

Would it be accurate to say that

 Sroot ksp - Sroot Q = 0.00105 M/L is the amount precipitated?

Title: Re: Solubility, How much precipitates?
Post by: rabolisk on March 19, 2011, 01:59:08 AM

No. You have to do the ICE table to get the exact amount. Assuming you just have a simple salt of single charged cation and anion, then you have X⁺ + Y^-  ::equil:: XY

Q = [X₀][Y₀]
Ksp = [X][Y] at equilibrium
Q > Ksp

From this point on, it's no different from any other equilibrium problems.