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

Problem 15.41

What is the pOH of each of the following aqueous solutions?

(a) 0.073 M LiOH

(c) 0.045 M Ba(OH)2

(b) 1.75 M NaOH

(d) 9.1×10–2 M HClO4






Answer:

(a) LiOH(aq) Li+(aq) + OH(aq) (strong base)

0.073 M LiOH = 0.073 M OH

pOH = –log[OH] = –log[0.073] = 1.14



(b) NaOH(aq) Na+(aq) + OH(aq) (strong base)

1.75 M NaOH = 1.75 M OH

pOH = –log[OH] = –log[1.75] = 0.243



(c) Ba(OH)2(aq) Ba2+(aq) + 2 OH(aq) (strong base)

0.045 M Ba(OH)2 = 2(0.045) M OH = 0.090 M OH

pOH = –log[OH] = –log[0.090] = 1.046



(d) HClO4(aq) + H2O(l) H3O+(aq) + ClO4(aq) (strong acid)

9.1×10–2 M HClO4 = 9.1×10–2 M H3O+

pH = –log[H3O+] = –log[9.1×10–2] = 1.041

pOH = 14.00 – pH (assuming 25 oC) = 14.00 – 1.041 = 12.96


Alternatively: [OH] = Kw/[ H3O+] = 1.0×10–14/9.1×10–2 = 1.1×10–13

pOH = –log[OH] = –log[1.1×10–13] = 12.96