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Problem 15.61

For a solution that is 0.15 M H_{3}PO_{4}(*aq*), determine (**a**) pH,
(**b**) [H_{3}PO_{4}], (**c**) [H_{2}PO_{4}^{–}], (**d**) [HPO_{4}^{2–}], and (**e**) [PO_{4}^{3–}].

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

H |
+ |
H |
H |
+ |
H |
||

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 [H_{3}O^{+}] =
[H_{2}PO_{4}^{–}] = 0.029 M

[H_{3}PO_{4}] = 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:

H |
+ |
H |
H |
+ |
HPO |
||

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} = 4.6×10^{5}, YES!

x = [HPO_{4}^{2–}] = 6.3×10^{–8} M

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

HPO |
+ |
H |
H |
+ |
PO |
||

Initial |
6.3×10 |
0.029 |
0 |
||||

Change |
– x |
+ x |
+ x |
||||

Equilibrium |
6.3×10 |
0.029 + x |
x |

Approximate? 6.3×10^{–8}/4.2×10^{–13} = 1.5×10^{5}, YES!

x = [PO_{4}^{3–}] = 9.1×10^{–19} M

Then:

(

a) pH = –log[H_{3}O^{+}] = –log[0.029] = 1.54(

b) [H_{3}PO_{4}] = 0.12 M(

c) [H_{2}PO_{4}^{–}] = 0.029 M(

d) [HPO_{4}^{2–}] = K_{a2}= 6.3×10^{–8}M(

e) [PO_{4}^{3–}] = 9.1×10^{–19}M