Question

If the Ka of a monoprotic weak acid is 6.6×10−6, what is the pH of a 0.29 M solution of this acid?

The Ka of a monoprotic weak acid is 0.00604. What is the percent ionization of a 0.114 M solution of this acid?

Enough of a monoprotic weak acid is dissolved in water to produce a 0.0154 M solution. The pH of the resulting solution is 2.38. Calculate the K_a for the acid.

Enough of a monoprotic weak acid is dissolved in water to produce a 0.0186 M solution. The pH of the resulting solution is 2.60. Calculate the pKa for the acid.

Answer 1

1)
HA dissociates as:

HA          ----->     H+   + A-
0.29                 0         0
0.29-x               x         x


Ka = [H+][A-]/[HA]
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 ((6.6*10^-6)*0.29) = 1.383*10^-3

since c is much greater than x, our assumption is correct
so, x = 1.383*10^-3 M



So, [H+] = x = 1.383*10^-3 M


use:
pH = -log [H+]
= -log (1.383*10^-3)
= 2.859
Answer: 2.86

2)
HA dissociates as:

HA          ----->     H+   + A-
0.114                 0         0
0.114-x               x         x


Ka = [H+][A-]/[HA]
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 ((6.04*10^-3)*0.114) = 2.624*10^-2

since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Ka = x*x/(c-x)
6.04*10^-3 = x^2/(0.114-x)
6.886*10^-4 - 6.04*10^-3 *x = x^2
x^2 + 6.04*10^-3 *x-6.886*10^-4 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 6.04*10^-3
c = -6.886*10^-4

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 = 2.791*10^-3

roots are :
x = 2.339*10^-2 and x = -2.943*10^-2

since x can't be negative, the possible value of x is
x = 2.339*10^-2

So, [H+] = x = 2.339*10^-2 M


use:
pH = -log [H+]
= -log (2.339*10^-2)
= 1.6309
Answer: 1.63

3)

use:
pH = -log [H+]
2.38 = -log [H+]
[H+] = 4.169*10^-3 M
HA dissociates as:

HA          ----->     H+   + A-
1.54*10^-2                 0         0
1.54*10^-2-x               x         x


Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Ka = 4.169*10^-3*4.169*10^-3/(0.0154-4.169*10^-3)
Ka = 1.547*10^-3
Answer: 1.55*10^-3

4)
use:
pH = -log [H+]
2.6 = -log [H+]
[H+] = 2.512*10^-3 M
HA dissociates as:

HA          ----->     H+   + A-
1.86*10^-2                 0         0
1.86*10^-2-x               x         x


Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Ka = 2.512*10^-3*2.512*10^-3/(0.0186-2.512*10^-3)
Ka = 3.922*10^-4
Given:
Ka = 3.922*10^-4

use:
pKa = -log Ka
= -log (3.922*10^-4)
= 3.4065
Answer: 3.41