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50 mL acid is titrated against 0.25N bas...

50 mL acid is titrated against `0.25N` base. It took `22.3 mL` of base to reach end-point. What is the normality of the acid ?

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`A _("Acid") xx V _("Acid") = N_("Base") xx V _("Base")`
Normality of the acid `= N _("Acid" ) = ( N _("Base") xx V _("Base"))/( V _("Acid")) = N _("Acid") = ( 0. 25 xx 22. 3)/(50) = 0. 111 5`
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