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25.0 cm^(3) of an acid required exactly ...

`25.0 cm^(3)` of an acid required exactly `20.5 cm^(3)` of deci molar base for complete neutralization. What is the normality of the acid?

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`(V xx M)_("acid") = (V xx M)_("base")`
Molarity of acid `=(V xx M)_("base")/("volume of acid") = (28.4 cm^(3) xx 0.25 M)/(20.0 cm^(3)) = 0.3550` M .
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