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What volume of 2.5*10^-2 M H2SO4 is required to neutralize 125mL if 0.03 M KOH
7 years ago

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Answered by bobpursley
changing concentrations to Normality: 2.5e-2 M= 5e-2N on sulfuric
.03M KoH=.03N KOH

Na*Va=Nb*Vb
Va=(.03/5e-2)125ml=75ml
7 years ago

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