Calculate the pH of a solution made by dissolving 10.6 g caffeine

[a.a]

1.

Homework Statement

Calculate the pH of a solution made by dissolving 10.6 g caffeine (C8H10N4O2, a weak base, Kb = 4.1 x 10-4) in enough water to make 100. mL of solution.

Homework Equations

pH = -log [H3O+]
kb = kw/ka

The Attempt at a Solution

I first found conc.:
MM= 194.2
m = 10.6
n = 0.05458 mol
conc. = 0.5458

kb = (10^-14)/ (4.1 x 10-4)
= 2.439*10^11

kb= x^2 / 0.5458
... x = 3.64 *10^-6
pOH = -log pOH
pH = 14 - pOH
= 8.56

But this is wrong... where did I go wrong?

 

[Borek]

kb = (10^-14)/ (4.1 x 10-4)

You were already given Kb.

 

[Blueshift5]

I would like to point out that the value for Kb you have used is incorrect. The correct value for Kb of caffeine is 4.1 x 10^-4, not 4.1 x 10^-11 as used in your calculation. This could be the reason for your incorrect result.

Additionally, it is important to note that the concentration of caffeine in the solution is not the same as the concentration of the base (C8H10N4O2) in the solution. The concentration of the base can be calculated by dividing the moles of caffeine by the volume of the solution (100 mL), which would give a concentration of 0.005458 M.

Using the correct Kb value and the concentration of the base, the calculation would be as follows:

Kb = 4.1 x 10^-4
[OH-] = √(Kb * [base]) = √(4.1 x 10^-4 * 0.005458) = 1.25 x 10^-3
pOH = -log [OH-] = -log (1.25 x 10^-3) = 2.90
pH = 14 - pOH = 14 - 2.90 = 11.10

Therefore, the pH of the solution made by dissolving 10.6 g of caffeine in 100 mL of water is 11.10. I hope this helps clarify any confusion and leads to the correct solution.