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 570 mg of acetylsalicylic acid each, in 300 mL of water?

Answer 1

Molar mass of HC9H7O4 = 7*MM(H) + 9*MM(C) + 4*MM(O)

= 7*1.008 + 9*12.01 + 4*16.0

= 179.146 g/mol

mass of HC9H7O4 = 2*570 mg

= 1140 mg

= 1.140 g [using conversion 1 g = 1000 mg]

we have below equation to be used:

number of mol of HC9H7O4,

n = mass of HC9H7O4/molar mass of HC9H7O4

=(1.14 g)/(179.146 g/mol)

= 6.364*10^-3 mol

volume , V = 300 mL

= 0.3 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 6.364*10^-3/0.3

= 2.121*10^-2 M

Lets write the dissociation equation of HC9H7O4

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

2.121*10^-2 0 0

2.121*10^-2-x x x

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

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that 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)*2.121*10^-2) = 2.646*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/(2.121*10^-2-x)

6.999*10^-6 - 3.3*10^-4 *x = x^2

x^2 + 3.3*10^-4 *x-6.999*10^-6 = 0

Let's solve this quadratic equation

Comparing it with general form: (ax^2+bx+c=0)

a = 1

b = 3.3*10^-4

c = -6.999*10^-6

solution of quadratic equation is found by below formula

x = {-b + √(b^2-4*a*c)}/2a

x = {-b - √(b^2-4*a*c)}/2a

b^2-4*a*c = 2.811*10^-5

putting value of d, solution can be written as:

x = {-3.3*10^-4 + √(2.811*10^-5)}/2

x = {-3.3*10^-4 - √(2.811*10^-5)}/2

solutions are :

x = 2.486*10^-3 and x = -2.816*10^-3

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

x = 2.486*10^-3

so.[H+] = x = 2.486*10^-3 M

we have below equation to be used:

pH = -log [H+]

= -log (2.486*10^-3)

= 2.60

Answer: 2.60