What is the pH of a 0.0538 M solution of Sr(OH)2?

asked Mar 13, 2021 95.3k views

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Answer:

pH= 13.03

Step-by-step explanation:

Since dissociation of Sr(OH)2= Sr2+ + 2[OH-]

[OH-]=2×0.0538 = 0.1076M

pOH= -log[0.1076]= 0.97

pH= 14-pOH = 14-0.97= 13.03

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