idkanymore33 idkanymore33

14-04-2020
Chemistry

contestada

Calculate the pH of a 0.025 M Ba(OH)2 solution

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averybutler8 averybutler8

14-04-2020

Ba(OH)2 --> Ba^2+ + 2OH-
Ba(OH)2: 0.025M
OH-: 2 x 0.025 = 0.050M = 5.0 x 10^-2M

pOH = - log [OH-] = - log (5.0 x 10^-2) = 2 - log 5 = 1.3

pH + pOH = 14; pH = 14 - 1.3 = 12.7

Ans: pH = 12.7

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