##### 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–].

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!

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