Textbook Problem 6B.5, Part C

[Myracle Franco 2G]

Hi! I was checking my answer for 6B.5, and I am unsure why my answer was different from the textbook. For Part C I did, -log(0.0092) and got about 2.04 as the PH and then did 14-2.04 = 11.96 for POH. However, both these answers are incorrect. What step am I missing? The textbook says the answers are PH: 1.74 and POH: 12.26.

Any feedback is appreciated, thanks!

[chloe lee]

Hi, it's because you need to multiple 0.0092 M for the concentration of OH-. This is because there are 2 moles of OH- in Ba(OH)2.

2 x 0.0092 M = 0.0184 M OH-
-log(0.0184) = 1.74
thus pOH = 1.74
14 - 1.74 = 12.26
thus pH = 12.26

hope this helps.