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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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`N_("acid") xx V_("acid") = N_("base") xx V_("base")`
Normality of the acid `=N_("acid") = (N_("base") xx V_("base"))/(V_("acid"))`
`=N_("acid")= (0.25xx22.3)/(50)=0.1115`
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