the much used drug cocoaine is an alkaloid. Alkaloids are noted for their bitter taste, and indication of their basic properties. Cocaine, C17H21NO4 is soluble in water tot he extent of 0.17 g/100 ml solution, and a saturated solution has a pH of 10.08. What is the Kb for cocaine?
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.
The answer is 8.5x10^-9
To find the value of Kb (base dissociation constant) for cocaine, we need to use the given information about its solubility in water and the pH of a saturated solution.
Here's how we can calculate the Kb:
1. Start by writing the balanced chemical equation for the dissociation of cocaine in water:
Cocaine (C17H21NO4) + H2O ⇌ Cocaine anion (C17H21NO4-) + H3O+
2. From the equation, we can see that cocaine (C17H21NO4) acts as an acid, donating a proton (H+) to water (H2O) to form H3O+ (hydronium ion) and the cocaine anion (C17H21NO4-).
3. Now, we must determine the concentration of the cocaine anion (C17H21NO4-) in the saturated solution. Since the pH of the solution is given, we know the concentration of H3O+ (hydronium ion) is 10^(-10.08) M, as pH = -log([H3O+]).
4. Since cocaine is still a weak base, we assume that most of it remains undissociated, so the concentration of cocaine (C17H21NO4) can be approximated as the initial concentration.
5. Using the solubility information given, we know that 0.17 g of cocaine dissolves in 100 mL of water. To convert this into molar concentration, we need to calculate the molar mass of cocaine:
C (12.01 g/mol) x 17 + H (1.01 g/mol) x 21 + N (14.01 g/mol) + O (16.00 g/mol) x 4 = 303.34 g/mol
So, 0.17 g is equal to 0.17/303.34 mol.
The concentration (C) of cocaine is then (0.17/303.34) mol/0.1 L (100 mL = 0.1 L).
6. Apply the equation for Kb:
Kb = [Cocaine anion][H3O+]/[Cocaine]
Substituting the values we calculated:
Kb = (10^(-10.08) M) / (0.17/303.34 M)
Simplifying:
Kb = (10^(-10.08) M) x (303.34 M) / 0.17
Calculating the value of Kb gives:
Kb ≈ 2.61 x 10^(-6) (rounded to four significant figures)
Therefore, the Kb value for cocaine is approximately 2.61 x 10^(-6).