Question

A 0.21 M solution of a weak base A‑ is made. To a limited extent, the A‑ reacts with H2O to form some OH- and some of the corresponding weak acid HA. If the weak acid has a pKa of 9.4, what will be the pH of the solution (to the nearest hundredths)? For a hint see the Acids and Bases handout on Canvas.

Answer 1

use:

pKa = -log Ka

9.4 = -log Ka

Ka = 3.981*10^-10

find kb for A-

use:

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/3.981*10^-10

Kb = 2.512*10^-5

A- dissociates as

A- + H2O -----> HA + OH-

0.21 0 0

0.21-x x x

Kb = [HA][OH-]/[A-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((2.512*10^-5)*0.21) = 2.297*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Kb = x*x/(c-x)

2.512*10^-5 = x^2/(0.21-x)

5.275*10^-6 - 2.512*10^-5 *x = x^2

x^2 + 2.512*10^-5 *x-5.275*10^-6 = 0

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

a = 1

b = 2.512*10^-5

c = -5.275*10^-6

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 = 2.11*10^-5

roots are :

x = 2.284*10^-3 and x = -2.309*10^-3

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

x = 2.284*10^-3

use:

pOH = -log [OH-]

= -log (2.284*10^-3)

= 2.64

use:

PH = 14 - pOH

= 14 - 2.64

= 11.36