Na_2CO_3 (molar mass 106 g/mol) reacts with excess HCl to give CO2 and water. Its equivalent mass in this reaction is:

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Na $_{2}$ CO $_{3}$ (molar mass 106 g/mol) reacts with excess HCl to give $CO X_{2}$ and water. Its equivalent mass in this reaction is:

CO X_{3} X^{2 -}

accepts 2

H X^{+}

to form

H X_{2} CO X_{3}

(

CO X_{2} + H X_{2} O

), so n-factor = 2. Equivalent mass = 106/2 = 53 g/eq.