Question

2. The pk of caffeine is 8.80 at 25°C. On average, a 12.0 oz cup of coffee contains 143 mg of caffeine. I prefer my coffee to be room temperature because of a very traumatic incident involving McDonald'soem coffee, so I brew an average cup of coffee at 25°C. My head hurts from studying for chemistry, so I drop in two extra strength tablets of aspirin, each containing 500 mg of acetylsalicylic acid. Assuming all the aspirin is dissolved, what are the equilibrium concentrations of caffeine, acetylsalicylic acid, and their conjugates at equilibrium? Bonus: What are the equilibrium concentrations of H30+ and OH?

Answer 1

Here the given caffeine is a weak base. Lets take it as (HB)

Now caffine can aceept H+ to form its conujugate acid (H2B+) as

HB + H2O -----------> H2B+ + OH-  

Here the base disscociation constant (Kb) is

Kb = [H2B+] * [OH-] / [HB]

Again Kb can be calculated from pKb as-

Kb = 10-pKb = 10-8.80 = 1.58 * 10-9  

Now given volume of caffine taken = 12 oz

= 12 * 29.5 mL (1 Oz = 29.5 mL)

= 354‬ mL

= 0.354 L

Given mass of caffine present = 143 mg

= 0.143 g

So molees of caffine present = mass / molar mass of caffine

= 0.143 g / 194.19 g/mol

= 7.36 * 10-4 moles

That means initial concentration of caffine present = moles / volume'

= 7.36 * 10-4 moles / 0.354 L

= 0.0021 moles/L

= 0.0021 M

Now to find the concentration of caffine and its conjugate acid , lets form the ICE table-

Reaction	HB + H2O ----------->	H2B+ +	OH-
Initial	0.0021 M	0	0
Change	-x	+x	+x
Equilibrium	0.0021 - xM	x	x

Putting the values-

Kb = [H2B+] * [OH-] / [HB]

1.58 * 10-9 = [x] * [x] / [0.0021 - x]

(1.58 * 10-9) * [0.0021 - x] = x2

(0.0033 * 10-9) - (1.58 * 10-9)x = x2

x2 +  (1.58 * 10-9)x - (0.0033 * 10-9) = 0

Solving this-

x = 1.8 * 10-6

So we can say, before addition of acetylsalicylic acid, the

[HB] = 0.0021 - x = 0.0021 - 1.8 * 10-6 = 0.00209 M

[H2B+] = x = 1.8 * 10-6 M

Now given mass of acetylsalicylic acid added = 500 mg = 0.5 g

So moles of  acetylsalicylic acid added = mass / molar mass of  acetylsalicylic acid

= 0.5 g / 180.158 g/mol

= 0.0028 moles

So concentration of acetylsalicylic acid added = moles / volume

= 0.0028 moles / 0.354 L

= 0.0079 M

Now if we take acetylsalicylic acid as (HA), then it will dissociate as-

HA ----------> H+ + A- (conjugate bse)

So from the dissociation of 0.0156 M, we will get 0.0079 M of H+ and 0.0079 M of A-

Again this H+ can react with HB to form H2B+ as

H+ + HB ----------> H2B+

That means we can say, due to addition of  acetylsalicylic acid as (HA), the amount of caffine (HB) from the solution will decrease and the amount of the conjugate base of caffine (H2B+) will increase.

So the final concentrations wiill be-

Reaction	H+ (from acetylsalicylic acid) +	HB ---------->	H2B+
Initial	0.0079 M	0.00209 M	1.8 * 10-6 M
Change	-0.00209	-0.00209	+0.00209
Equilibrium	0.00581 M	0 M	0.0020918‬ M

That means we can say, after this point

[acetylsalicylic acid] = [H+] = 0.00581 M

[conjugate base of acetylsalicylic acid] = [A-] = 0.0079 M

[caffine] = [HB] = 0

[Conjugate acid of caffine] = [H2B+] = 0.0020918‬ M