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What is the pH of 0.05 m baoh2? |
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Answer» tion:Ba(OH)2 ⟶Ba + 2 +2OH − ∴[OH − ] = 0.05 × 2 = 0.10M⇒POH = −log(0.10) = 1As we know that,PH + pOH = 14∴pH=14−pOH=14−1=13 |
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