Question

An aqueous solution of formic acid, HCOOH, is prepared at 0.075 M. The acid-ionization constant is K_a = 1.8 x 10^-4. Because this is a weak acid, you may presume that the ionic strength is negligible (m = 0). Calculate:

a) the pK_a of HCOOH

b) [H⁺]

c) [HCOO^-]  

d) [HCOOH] at equilibrium

Answer 1

a)

Given:

Ka = 1.8*10^-4

use:

pKa = -log Ka

= -log (1.8*10^-4)

= 3.7447

Answer: 3.74

b)

HCOOH dissociates as:

HCOOH -----> H+ + HCOO-

7.5*10^-2 0 0

7.5*10^-2-x x x

Ka = [H+][HCOO-]/[HCOOH]

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)*7.5*10^-2) = 3.674*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/(7.5*10^-2-x)

1.35*10^-5 - 1.8*10^-4 *x = x^2

x^2 + 1.8*10^-4 *x-1.35*10^-5 = 0

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

a = 1

b = 1.8*10^-4

c = -1.35*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 = 5.403*10^-5

roots are :

x = 3.585*10^-3 and x = -3.765*10^-3

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

x = 3.585*10^-3

So, [H+] = x = 3.585*10^-3 M

[HCOO-] = x = 3.585*10^-3 M

[HCOOH] = 0.075-x

= 0.075 - 3.585*10^-3

= 0.0714 M

b) 3.58*10^-3 M

c) 3.58*10^-3 M

d) 0.071 M