A 0.0133mol sample of Ba(OH)2 is dissolved in water to make up 1.5L of solution. What is the pH of the solution at 25.0∘C? ?

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  • Bobby
    Lv 7
    2 months ago

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

    moles of Ba(OH)2 = 0.0133mol 

    moles of OH- = 2* 0.0133mol  = 0.0266 moles 

    in 1.5 L [OH-] = 0.0266 moles / 1.5L = 0.0177 M

    pOH = 1.75

    pH = 12.2

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