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ТаняTanya07
@ТаняTanya07
July 2022
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Рассчитать pH 0,02М раствора муравьиной кислоты,если Kа=1,86×10^-4
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janislav
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PH=-log( √(Ka*C(HCOOH) )) = -log( √(1,86×10∧-4*0,02) )= 2,71
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Verified answer
PH=-log( √(Ka*C(HCOOH) )) = -log( √(1,86×10∧-4*0,02) )= 2,71