Question

1. Determine the pH for the following Show all work and please be neat

A. (3 points) 0.0055 M Ba(OH)₂ solution

          

             

B. (3 points) 0.040 M H₂SO₄ solution

C. (7 points) 0.27 M HNO₂ (HNO₂, K_a = 4.0  10^–4)

             

D. (7 points) 0.39 M NH₄Cl (pK_b for NH₃ = 4.74)           

Answer 1

A)

[OH-] = 2*[Ba(OH)2] = 2*0.0055 = 1.1*10^-2 M

use:

pOH = -log [OH-]

= -log (1.1*10^-2)

= 1.96

use:

PH = 14 - pOH

= 14 - 1.96

= 12.04

Answer: 12.04

B)

[H+] = 2*[H2SO4] = 2*0.040 = 0.080 M

use:

pH = -log [H+]

= -log (8*10^-2)

= 1.10

Answer: 1.10

C)

HNO2 dissociates as:

HNO2 -----> H+ + NO2-

0.27 0 0

0.27-x x x

Ka = [H+][NO2-]/[HNO2]

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 ((4*10^-4)*0.27) = 1.039*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)

4*10^-4 = x^2/(0.27-x)

1.08*10^-4 - 4*10^-4 *x = x^2

x^2 + 4*10^-4 *x-1.08*10^-4 = 0

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

a = 1

b = 4*10^-4

c = -1.08*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 = 4.322*10^-4

roots are :

x = 1.019*10^-2 and x = -1.059*10^-2

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

x = 1.019*10^-2

so.[H+] = x = 1.019*10^-2 M

use:

pH = -log [H+]

= -log (1.019*10^-2)

= 1.99

Answer: 1.99

D)

find Ka for NH4+

use:

pKb = -log Kb

4.74-log Kb

Kb = 1.82*10^-5

use:

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

Ka = (1.0*10^-14)/1.82*10^-5

Ka = 5.495*10^-10

NH4+ + H2O -----> NH3 + H+

0.39 0 0

0.39-x x x

Ka = [H+][NH3]/[NH4+]

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 ((5.495*10^-10)*0.39) = 1.464*10^-5

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

so, x = 1.464*10^-5 M

use:

pH = -log [H+]

= -log (1.464*10^-5)

= 4.83

Answer: 4.83