Monday
July 28, 2014

Homework Help: chemistry

Posted by For Dr. Bob on Sunday, November 15, 2009 at 10:24pm.

The question asks, "Butanoic acid has a partition coeff of 3.0 when distributed b/t water and benzene. Find the formal concentration of butanoic acid in each phase when 100 mL of 0.10 M aqueous butanoic acid is extracted with 25 mL of benzene at pH 10.00."

I know that the first step is to find D, where D = K(Ka)/[(Ka+[H+)].

So I calculated:
D = 3.0(1.52x10^-5)/[(1.52x10^-5+10^-10)].

The answer should have been D = 1.97 x 10^-5, but I didn't get this.

You answered: If the question is worded correctly, then Kd water/benzene = 3.0
However, the distribution coefficient USUALLY is listed as organic/water.

So my question is, what do I need to do to get 1.97 x 10^-5? Where did I go wrong?

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