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`{:(2NaOH+H_(2)SO_(4)rarrNa_(2)SO_(4)+2H_(2)O),("Base Acid"):}` write the valency factor of NaOH.

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Valency factor of base = 1
Here, one of molecule of `H_(2)SO_(4)` replaced one `OH^(-)` from NaOH. Therefore, valency factor for `H_(2)SO_(4)` is one
`:. ` Eq. wt. of `H_(2)SO_(4)=("Mol.wt")/(1)`

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