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Calculate [OH^(-)] if pOH = 8.3. |
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Answer» SOLUTION :`pOH=-log_(10)[OH^(-)],8.3=-log_(10)[OH^(-)]` TAKING antilog on both the sides `[OH^(-)]` = anti LOG(-8.3), `[OH^(-1)]` = antilog(-9-8.3+9) = antilog`(0.7)xx10^(-9)` = `5.012xx10^(-9)mol//dm^(3)`. |
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