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

Problem 15.54

Piperidine, C5H11N (pKb = 2.89), is a colorless liquid having the odor of pepper. Calculate the pH of 0.00250 M C5H11N.






Answer:

Treat this as any equilibrium problem: write the balanced chemical reaction, write the mass action expression for Kb, set up a table of concentrations, and then solve for the unknown variables using the given information. Assume 25 oC.

C5H11N(aq)
+
H2O(l)
[C5H11NH]+(aq)
+
OH(aq)
Initial
0.00250
0
0
Change
– x
+ x
+ x
Equilibrium
0.00250 – x
x
x

Approximate? 0.00250/1.3×10–3 = 2 < 100 No!

Solve this using the quadratic equation to give:

x = 1.3×10–3, –2.6×10–3

Only the positive root makes sense, so [OH] = x = 1.3×10–3 M

pOH = –log[OH] = –log(1.3×10–3) = 2.89

pH = 14.00 – pOH = 14.00 – 2.89 = 11.11