Question

a) Calculate the pH of a 0.250 M solution of formic acid, HCO₂H?  

     K_a, HCO₂H = 1.8 x 10^-4.   HCO₂H(aq)  ⇌HCO₂^-(aq) + H⁺(aq)

b) What is the concentration of HCO₂H at equilibrium?

Answer 1

a)

HCO2H dissociates as:

HCO2H -----> H+ + HCO2-

0.25 0 0

0.25-x x x

Ka = [H+][HCO2-]/[HCO2H]

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 ((1.8*10^-4)*0.25) = 6.708*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)

1.8*10^-4 = x^2/(0.25-x)

4.5*10^-5 - 1.8*10^-4 *x = x^2

x^2 + 1.8*10^-4 *x-4.5*10^-5 = 0

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

a = 1

b = 1.8*10^-4

c = -4.5*10^-5

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 = 1.8*10^-4

roots are :

x = 6.619*10^-3 and x = -6.799*10^-3

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

x = 6.619*10^-3

So, [H+] = x = 6.619*10^-3 M

use:

pH = -log [H+]

= -log (6.619*10^-3)

= 2.1792

Answer: 2.18

b)

[HCO2H] = 0.25-x

= 0.25 - 6.619*10^-3

= 0.243 M