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

Problem 15.109

Calculate the pH of a solution that is 1.0×10–8 M HCl.






Answer:

Both water and HCl contribute significant amounts of hydronium ion, so both reactions must be considered. This problem, then, is a common–ion problem where HCl provides the common ion to the water autoionization equilibrium. So, write the two reactions, write the mass action expression for Kw, set up a table of concentrations, and solve. Assume 25 oC.
HCl(aq)
+
H2O(l)
H3O+(aq)
+
Cl(aq)
2H2O(l)
H3O+(aq)
+
OH(aq)
Kw = [H3O+]e[OH]e = 1.0×10–14
Initial
1.0×10–8
0
Change
+ x
+ x
Equilibrium
1.0×10–8 + x
x

1.0×10–14 = [1.0×10–8 + x][x]

1.0×10–14 = 1.0×10–8x + x2

x2 + 1.0×10–8x – 1.0×10–14 = 0

Solve with the quadratic equation to give

x = 9.5×10–8, –1.1×10–7

Only the positive value makes sense chemically, so

[H3O+] = 1.0×10–8 + x = 1.0×10–8 + 9.5×10–8 = 1.05×10–7 M

pH = –log[H3O+] = –log[1.05×10–7] = 6.979 = 6.98 (despite the sig fig rules, pH values beyond 2 decimal places are nearly always meaningless)