pH + pOH = 14
pOH = 14-10.08 = 3.92
[OH-]=-log^-1(3.92)=1.2002x10^-4
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Kb = [BH+][OH-] / [B], with B = base
here,
Kb = [C17H21O4NH+][OH-] / [C17H21O4N]
since [C17H21O4NH+]=[OH-]
Kb = [OH-]^2 / {[C17H21O4N]-[OH-]}
Since [OH-] is much smaller than the overall concentration of the base, we can approximate:
Kb = [OH-]^2 / [C17H21O4N]
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Find the molarity of C17H21O4N and substitute it into the formula for Kb along with the value of [OH-].
The much-abused drug cocaine is an alkaloid. Alkaloids are noted for their bitter taste, an indication their basic properties. Cocaine C17H21O4N, is soluble in water the the extent of 0.17g/100 mL solution, and a saturated solution has a PH = 10.08. What is the value of Kb for cocaine?
C17H21O4N + H2O --> <--- C17H21O4NH+ + OH- Kb = ?
I would appreciate if someone would do this problem without skipping any steps, I need to understand it completely for an exam. With the data 0.17g/100 mL how do I find M (concentration)? Thanks.
5 answers
If the solubility is 0.17 g/100 mL, that is equivalent to 1.7 g/1000 mL. And moles = grams/molar mass will give you the molarity.
Use the equation given.
You know pH, convert that to pOH, then to OH^-.
Kb = (C17H21O4NHH^+)(OH^-)/(C17H21O4N).
You know OH^- now, that is the same as C17H21O4NH^+, and the C17H21O4N is 1.7/molar mass - (OH^-). You may be able to neglect OH^- in the denominator if it is small in comparison to the cocaine molarity. Post your work if you get stuck.
Use the equation given.
You know pH, convert that to pOH, then to OH^-.
Kb = (C17H21O4NHH^+)(OH^-)/(C17H21O4N).
You know OH^- now, that is the same as C17H21O4NH^+, and the C17H21O4N is 1.7/molar mass - (OH^-). You may be able to neglect OH^- in the denominator if it is small in comparison to the cocaine molarity. Post your work if you get stuck.
kb=2,6 x 10^-6
is it correct?
is it correct?
No
Is 3.3*10^-9