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

Problem 15.51

A 1.00–g sample of aspirin (acetylsalicylic acid) is dissolved in 0.300 L of water at 25 oC, and its pH is found to be 2.62. What is the Ka of aspirin?

o–C6H4(OCOCH3)COOH(aq) + H2O(l)H3O+(aq) + o–C6H4(OCOCH3)COO(aq) Ka = ?






Answer:

Find the molarity of the aspirin sample, use the balanced reaction, write the mass action expression for Ka, set up a table of concentrations, and solve.

o–C6H4(OCOCH3)COOH = C9H8O4 = 9(12.011) + 8(1.008) + 4(15.999) = 180.159 g/mol

initial molarity = (1.00 g/180.159 g/mol)/0.300 L = 0.0185 M


o–C6H4(OCOCH3)COOH(aq)
+
H2O(l)
H3O+(aq)
+
o–C6H4(OCOCH3)COO(aq)
Initial
0.0185
0
0
Change
– x
+ x
+ x
Equilibrium
0.0185 – x
x
x

Since Ka is unknown, x must be found from other information, in this case the pH.

x = [H3O+]e = 10–pH = 10–2.62 = 2.4×10–3 M = [C9H7O4]e

[C9H8O4]e = 0.0185 – 2.4×10–3 = 0.0161 M

These concentrations can be used to find Ka