Question

A. if the Ka of a monoprotic weak acid is 1.1 x 10^-6, what is the ph of a 0.21M solution of this acid?

ph=

B. Enough of a monoprotic acid is dissolved in water to produce a 0.0117M solution. The PH of the resulting solution is 2.57. Calculate the Ka for this acid.

Ka=

C.The Ka of a monoprotic weak acid is 7.73x10 ^-3. what is the percent ionization of a 0.106 M solution of this acid?

Answer 1

A)

HA dissociates as:

HA -----> H+ + A-

0.21 0 0

0.21-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 ((1.1*10^-6)*0.21) = 4.806*10^-4

since c is much greater than x, our assumption is correct

so, x = 4.806*10^-4 M

so.[H+] = x = 4.806*10^-4 M

use:

pH = -log [H+]

= -log (4.806*10^-4)

= 3.32

Answer: 3.32

B)

use:

pH = -log [H3O+]

2.57 = -log [H3O+]

[H3O+] = 2.692*10^-3 M

HA dissociates as:

HA -----> H+ + A-

1.17*10^-2 0 0

1.17*10^-2-x x x

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

Ka = 2.692*10^-3*2.692*10^-3/(0.0117-2.692*10^-3)

Ka = 8.04*10^-4

Answer: 8.04*10^-4

C)

HA dissociates as:

HA -----> H+ + A-

0.106 0 0

0.106-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 ((7.73*10^-3)*0.106) = 2.862*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)

7.73*10^-3 = x^2/(0.106-x)

8.194*10^-4 - 7.73*10^-3 *x = x^2

x^2 + 7.73*10^-3 *x-8.194*10^-4 = 0

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

a = 1

b = 7.73*10^-3

c = -8.194*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 = 3.337*10^-3

roots are :

x = 2.502*10^-2 and x = -3.275*10^-2

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

x = 2.502*10^-2

% dissociation = (x*100)/c

= 2.502*10^-2*100/0.106

= 23.6 %

Answer: 23.6 %