Sr(OH)2 ----> Sr2+ +2OH-
This is the ionization of Sr(OH)2 in water. It shows 1mole
Sr(OH)2 produce 2 mole
OH-
therefore 0.0025 M Sr(OH)2 produce 2* 0.0025 M OH-
thus [OH] = 0.005 M
Now calculate pOH = - log[OH-] = 2.3
we have pH + pOH = 14
therefore pH = 14-2.3 = 11.7
pH = 11.7
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