Succinic acid is a diprotic acid with molar mass of 118 g/mol. calculate the mass of the acid required to neutralize 25.0 ml of a 0.100 m naoh solution.

 Because it is diprotic acid, we can write like R(COOH)2 ( it can be written as H2A also)
R(COOH)2+2NaOH > R(COONa)2+2H2O

25.0 ml of a 0.100 m NaOH solution
25.0 ml=0.0250 L solution Of NaOH
 0.100 m NaOH=0.100 mol/L
0.0250 L *0.100 mol/L=0.00250 mol NaOH
0.00250 mol NaOH*1mol R(COOH)2/2 mol NaOH=0.00125 mol acid
0.00125 mol R(COOH)2*118g R(COOH)2/1mol R(COOH)2 = 0.148 g acid

Answer is 0.148 g Succinic acid

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