CHM112 Home Page Reference Materials Homework Assignments Lectures Exams & Quizzes Grades Study Aids URI Home Page


J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 15.61

For a solution that is 0.15 M H3PO4(aq), determine (a) pH, (b) [H3PO4], (c) [H2PO4], (d) [HPO42–], and (e) [PO43–].






Answer:

This is a polyprotic weak acid. The first ionization is treated exactly like every other weak acid:

 

H3PO4(aq)

+

H2O(l)

H3O+(aq)

+

H2PO4(aq)

 

Initial

0.15

     

0

 

0

Change

– x

     

+ x

 

+ x

Equilibrium

0.15 – x

     

x

 

x


Approximate? 0.15/7.1×10–3 = 21, NO!

Using the quadratic equation gives

x = 0.029 or x = –0.036

Only the positive value makes sense, so [H3O+] = [H2PO4] = 0.029 M

[H3PO4] = 0.15 – 0.029 = 0.12 M


The second and third ionizations do not contribute significantly to these concentrations.


To find the hydrogen phosphate concentration, look at the second ionization step:

 

H2PO4(aq)

+

H2O(l)

H3O+(aq)

+

HPO42–(aq)

 

Initial

0.029

     

0.029

 

0

Change

– x

     

+ x

 

+ x

Equilibrium

0.029 –x

     

0.029 + x

 

x

Approximate? 0.029/6.3×10–8 = 4.6×105, YES!

x = [HPO42–] = 6.3×10–8 M

To find the phosphate concentration, look at the third ionization step:

 

HPO42–(aq)

+

H2O(l)

H3O+(aq)

+

PO43–(aq)

 

Initial

6.3×10–8

     

0.029

 

0

Change

– x

     

+ x

 

+ x

Equilibrium

6.3×10–8 –x

     

0.029 + x

 

x

Approximate? 6.3×10–8/4.2×10–13 = 1.5×105, YES!

x = [PO43–] = 9.1×10–19 M


Then:

(a) pH = –log[H3O+] = –log[0.029] = 1.54

(b) [H3PO4] = 0.12 M

(c) [H2PO4] = 0.029 M

(d) [HPO42–] = Ka2 = 6.3×10–8 M

(e) [PO43–] = 9.1×10–19 M