Answer:
Explanation:
Caffeine is a weak base with pKb = 10.4
Kb = 10⁻¹⁰°⁴ = 3.98 x 10⁻¹¹
molecular weight of caffeine = 194.2
455 x 10⁻³ g / L = 455 x 10⁻³ / 194.2 moles / L
concentration of given solution a = 2.343 x 10⁻³ M
Let the caffeine be represented by B .
B + H₂O = BH + OH⁻
a - x x x
x² / ( a - x ) = Kb
x² / ( a - x ) = 3.98 x 10⁻¹¹
x is far less than a so a -x is almost equal to a
x² = 3.98 x 10⁻¹¹ x 2.343 x 10⁻³ = 9.32 x 10⁻¹⁴
x = 3.05 x 10⁻⁷
[ OH⁻ ] = 3.05 x 10⁻⁷
pOH = - log ( 3.05 x 10⁻⁷ )
= 7 - log 3.05
= 7 - 0.484 = 6.5
pH = 14 - 6.5 = 7.5
The pH of 455 mg/L of caffeine is 7.5
Using the formula;
Mass concentration = molar concentration × molar mass
Molar mass of caffeine = 194 g/mol
Mass concentration of caffeine = 455 mg/L
Molar concentration = Mass concentration/molar mass
Molar concentration = 455 × 10^-3g/L/194 g/mol
= 0.00235 M
Let Caffeine by depicted by the general formula BH
We can now set up the ICE table as follows;
:B + H2O ⇄ BH + OH^-
I 0.00235 0 0
C - x +x +x
E 0.00235 - x x x
Note that water is present in large excess
Again; pKb of caffeine =10.4
Kb = Antilog[-pKb]
Kb = Antilog [-10.4]
Kb = 3.98 × 10^-11
Kb = [BH] [OH^-]/[:B]
3.98 × 10^-11 = [x] [x]/[ 0.00235 - x ]
3.98 × 10^-11 [ 0.00235 - x ] = [x] [x]
9.4 × 10^-14 - 3.98 × 10^-11x = x^2
x^2 + 3.98 × 10^-11x - 9.4 × 10^-14 = 0
x = 3.1 × 10^-7 M
Recall [BH] = [OH^-] = 3.1 × 10^-7 M
Now;
pOH = - log [OH^-]
pOH = log [3.1 × 10^-7 M]
pOH = 6.5
But;
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 6.5
pH = 7.5
The pH of 455 mg/L of caffeine is 7.5
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Missing parts
Caffeine (C8H10N4O2) is a weak base with a pKb of 10.4. Calculate the pH of a solution containing a caffeine concentration of 455 mg/L.
