Posted by Symone on Sunday, December 5, 2010 at 8:49pm.
Let's call cocaine something simple so as to save all that typing, say R3N.
R3N + HOH ==> R3NH^+ + OH^-
Kb = (R3NH^+)(OH^-)/(R3N)
pH = 10.08, solve for pOH from
pH + pOH = pKw = 14
Then calculate OH^- from
pOH = -log(OH^-). That will also be (R3NH^)+.
For R3N, substitute the solubility in M = moles/L.
0.17g/100 mL = 1.7 g/L and 1.7 is how many moles. moles = grams/molar mass
Post your work if you get stuck.
i got 3.2 for my pOH and 6.31x10^-4 for my [OH^-]. after this i am stuck and not sure where to go.
You got what for pOH.
pH + pOH = 14
pOH = 14-pH = 14-10.08 and that isn't 3.2. When you obtain the correct value for OH^-, then substitute that for (OH^-) in the Kb expression. Substitute the same number of (R3NH^+). That leaves something to be substituted for (R3N) in the Kb expression.
The problem tells you the solubility of cocaine is 0.17 g/100 mL. That is the same as 1.7 g/L. How many moles is that?
1.7/molar mass cocaine = moles and that in 1L is M. Substitute that and solve for Kb.
Kb= .0876???
no.
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