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Calculate [OH–] if pOH = 8.3. |
Answer» pOH = -log10[OH–]; 8.3 = -log10[OH–] Taking antilog on both the sides [OH–] = antilog (-8.3) = [OH–] = antilog (-9 - 8.3 + 9) = antilog [(0.7) × 10-9] = 5.012 × 10-9 mol/dm3 |
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