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Calculate [H^(+)] if pOH = 9.23. |
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Answer» Solution :`pH+pOH=14""pH=14-pOH=14-9.23=4.77` `pH=-log_(10)[H^(+)]=-4.77=LOG[H^(+)]` Taking ANTILOG on both the sides `[H^(+)]` = ANTI log (-4.77) = anti log[+5 - 4.77 - 5] = anti log[0.23]`xx10^(-5)` = `1.698xx10^(-5)mol//dm^(3)`. |
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