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प्रश्न
Calculate the pH of 1.5 × 10−3 M solution of Ba(OH)2.
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उत्तर
\[\ce{\underset{1.5×10^{-3}M}{Ba(OH)2} -> Ba^{2+} + \underset{2×1.5×10^{-3}M}{2OH^-}}\]
[OH–] = 3 × 10−3 M
[∵ pH + pOH = 14]
pH = 14 – pOH
pH = 14 – (– log [OH–])
= 14 + log [OH–]
= 14 + log (3 × 10−3)
= 14 + log 3 + log 10−3
= 11 + 0.4771
pH = 11.48
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