If 0.2178 g of a diprotic solid acid is neutralized by 44.81 mL of 0.0953 M NaOH, calculate the molar mass of the solid acid

Answer 1

R(CO2H)2+2NaOH-->R(CO2Na)2+2H2O

Mu(NaOH)=44,81*10^(-3)*0,0953=4,27*10^… mol
Mu(acid)=4,27*10^(-3):2=0,002135 mol
If R=aliphatic saturated rest and there are not other substituents in C-chain,then R=CxH2x and:
0,2178=0,002135(90+12x+2)=0,1964+0,025…
x=(0,2178-0,1964):0,0256=0,84~1

That acid is malonic acid and its molar mass is 104