Question

39) A 40.0 mL sample of 0.300 M formic acid is titrated with 0.200 M NaOH. Calculate the initial pH before the titration is begun. (а) 1.93 b) 2.13 (c) 2.41 (d) 2.85 (e) 0.55

Answer 1

HCOOH dissociates as:

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

0.3 0 0

0.3-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)*0.3) = 7.348*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/(0.3-x)

5.4*10^-5 - 1.8*10^-4 *x = x^2

x^2 + 1.8*10^-4 *x-5.4*10^-5 = 0

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

a = 1

b = 1.8*10^-4

c = -5.4*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 = 2.16*10^-4

roots are :

x = 7.259*10^-3 and x = -7.439*10^-3

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

x = 7.259*10^-3

So, [H+] = x = 7.259*10^-3 M

use:

pH = -log [H+]

= -log (7.259*10^-3)

= 2.13

Answer: b