Page 70 - [B._MURPHY,_C._MURPHY,_B._HATHAWAY]_A_working_meth

54 Chapter 5
Solution:

1. Read the question carefully-notice that the solution is a buffer
+ constant pH.
2. Identify the species present: H,P04(aq, -acid; NaH2P04(aql-
salt.
3. (a) H3P04 + H20 + H30+ + H2PO;;
(b) NaH2P04 -+ Na+ + H2P0,.
4.

0.35 mol 0.25 mol in
in 1 dm3 of H3P04 1 dm3 of NaH2P04
Therefore in the buffer solution:
0.35 moles of H3P04 in 2 dm3 + 0.175 moles H3P04 in 1 dm3,
i.e. 0.175 M;
0.25 moles of NaH2P04 in 2 dm3 0.125 moles NaH2P04 in 1
dm3, i.e. 0.125 M.

5. H3P04 + H20 + H30f + HZPO,
Initial 0.175 0 0.125
Change -x +X +X
Final (0.175 - x) X (0.125 + x)

6. Ka= ([H2P0,][H30f])/[H3P04]= ((0.125 + ~)(~))/(0.175 - X)
= 7.5 x 10-3 ..... (t).
7. Assume x << 0.175 and x << 0.125.
8. Ka = (0.125 ~)/(0.175) = 7.5 x + x = 0.0105 M.
9. (0.0105/0.125)?40 = 8.4% and (0.0105/0.175)?40 = 6%. Since
both are >5%, the assumptions made were invalid, and a
quadratic equation (t), must now be solved:
x2 + (0.125)~ = (-7.5 x 10d3)x + 0.0013125
+ x2 + (0.1325)x-0.0013125 = 0.
a = 1; b = 0.1325; c = -0.0013125, with solution
-b& d-
x=
2a
+ x = 0.009257 or x = -0.141759. The latter negative value of
x is meaningless. Therefore x = [H30]+ = 0.009257 M.
10. pH = -loglo[H30+] = -10glo(O.O092587) = 2.033.

Answer:pH = 2.03