Question

A typical aspirin tablet contains 424 mg of aspirin (C9O4H8), a monoprotic acid with a Ka= 2.8x10^-4. A monoprotic acid donates a single proton. If you dissolve two aspirin tablets in a 300 mL glass of water, what is the pH of the solution?

Answer 1

Molar mass of C9O4H8,

MM = 9*MM(C) + 4*MM(O) + 8*MM(H)

= 9*12.01 + 4*16.0 + 8*1.008

= 180.154 g/mol

mass(C9O4H8)= mass of 2 tablets

= 2*424 mg

= 848 mg

= 0.848 g

number of mol of C9O4H8,

n = mass of C9O4H8/molar mass of C9O4H8

=(0.848 g)/(180.154 g/mol)

= 4.707*10^-3 mol

volume , V = 300 mL

= 0.300 L

Molarity,

M = number of mol / volume in L

= 4.707*10^-3/0.3

= 1.57*10^-2 M

C9O4H8 dissociates as:

C9O4H8 -----> H+ + C9O4H7-

1.57*10^-2 0 0

1.57*10^-2-x x x

Ka = [H+][C9O4H7-]/[C9O4H8]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((2.8*10^-4)*1.57*10^-2) = 2.097*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

2.8*10^-4 = x^2/(1.57*10^-2-x)

4.396*10^-6 - 2.8*10^-4 *x = x^2

x^2 + 2.8*10^-4 *x-4.396*10^-6 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 2.8*10^-4

c = -4.396*10^-6

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 1.766*10^-5

roots are :

x = 1.961*10^-3 and x = -2.241*10^-3

since x can't be negative, the possible value of x is

x = 1.961*10^-3

so.[H+] = x = 1.961*10^-3 M

we have below equation to be used

pH = -log [H+]

= -log (1.961*10^-3)

= 2.71

Answer: 2.71