Question

The active ingredient in aspirin is acetylsalicylic acid (HC9H7O4) , a monoprotic acid with Ka=3.3×10−4 at 25 ∘C .

What is the pH of a solution obtained by dissolving two extra-strength aspirin tablets, containing 510 mg of acetylsalicylic acid each, in 300 mL of water?

Answer 1

Molar mass of C9H8O4,

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

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

= 180.154 g/mol

mass(C9H8O4)= 2*510 mg

= 1020 mg

= 1.02 g

number of mol of C9H8O4,

n = mass of C9H8O4/molar mass of C9H8O4

=(1.02 g)/(180.154 g/mol)

= 5.662*10^-3 mol

volume , V = 300 mL

= 0.3 L

Molarity,

M = number of mol / volume in L

= 5.662*10^-3/0.3

= 0.01887 M

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HC9H7O4 dissociates as:

HC9H7O4 -----> H+ + C9H7O4-

1.887*10^-2 0 0

1.887*10^-2-x x x

Ka = [H+][C9H7O4-]/[HC9H7O4]

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 ((3.3*10^-4)*1.887*10^-2) = 2.495*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)

3.3*10^-4 = x^2/(1.887*10^-2-x)

6.227*10^-6 - 3.3*10^-4 *x = x^2

x^2 + 3.3*10^-4 *x-6.227*10^-6 = 0

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

a = 1

b = 3.3*10^-4

c = -6.227*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 = 2.502*10^-5

roots are :

x = 2.336*10^-3 and x = -2.666*10^-3

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

x = 2.336*10^-3

so.[H+] = x = 2.336*10^-3 M

use:

pH = -log [H+]

= -log (2.336*10^-3)

= 2.63

Answer: 2.63