What is the pH of a 0.235 M Ba(OH)2 solution?

13.672

Explanation:

2 moles of OH in Ba(OH)₂

0.235 M × 2 mol OH = 0.47 M OH

pOH = -log[OH-] ; pOH = -log[0.47 M OH] ; 0.328 = pOH (consider significant figures)

14 - pOH = pH because pH + pOH = 14

14 - 0.328 = 13.672 [there are three decimal places after the OH concentration, so there should be in the answer as well - 13.672]

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