## CHM 112 Section 1

Homework Assignment #13
##### Due March 1, 2006 by 8:05 a.m.

Practice (not to be turned in): Problems 15.39, 15.41, 15.44, 15.108, 15.109

1) A solution of perchloric acid is prepared that has a molar concentration equal to the last four digits of your Student ID Number divided by 107, rounded to two significant figures. What is the pH and pOH of that solution at 25 oC?

2) A solution of nitric acid is prepared that has a molar concentration equal to the last four digits of your Student ID Number divided by 105, rounded to two significant figures. What is the pH and pOH of that solution at 25 oC?

3) A solution of potassium hydroxide is prepared that has a molar concentration equal to the last four digits of your ID Number divided by 106, rounded to two significant figures. What is the pH and pOH of that solution at 25 oC?

4) A solution of barium hydroxide is prepared that has a molar concentration equal to the last four digits of your ID Number divided by 107, rounded to two significant figures. What is the pH and pOH of that solution at 25 oC? Assume complete solubility and complete ionization.

Be careful of significant figures.

1) For perchloric acid:

HClO4(aq) + H2O(l) H3O+(aq) + ClO4(aq)

So the [H3O+] = [HClO4] = ID/107 M

pH = –log[H3O+]

pOH = 14.00 – pH (at 25 oC)

2) For nitric acid:

HNO3(aq) + H2O(l) H3O+(aq) + NO3(aq)

So the [H3O+] = [HNO3] = ID/105 M

pH = –log[H3O+]

pOH = 14.00 – pH (at 25 oC)

3) For potassium hydroxide:

KOH(aq) K+(aq) + OH(aq)

So [OH] = [KOH] = ID/106 M

pOH = –log[OH]

pH = 14.00 – pOH (at 25 oC)

4) For barium hydroxide:

Ba(OH)2(aq) Ba2+(aq) + 2 OH(aq)

So [OH] = [Ba(OH)2] = 2×ID/107 M

pOH = –log[OH]

pH = 14.00 – pOH (at 25 oC)

Since all concentrations are rounded to 2 significant figures, all pH and pOH values are found to 2 places past the decimal.