Question

calculate the ph of 0.05m ba(oh)2 solution

Answer 1

[OH-] = N = 2×0.05 = 0.1
pOH = -log[OH-] = -log(0.1) = 1
PH = 14-pOH = 14-1 = 13

Answer 2

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SOLUTION:-

Ba(OH)2 - Ba2+ + 2OH-

0.05 M X 2= 0.10M pOH= -log(0.10 M) = 1.00

pH = 14-pOH= 14 - 1.00=13

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