How many Liters of a 0.1154 M HCl solution are required to neutralize 0.4075 grams of TRIS?

Answer 1

Tris has MW=121.14
It reacts 1:1 with HCl. Thus
mole HCl =mole Tris=>
M(HCl)*V(HCl) = mass(Tris) / MW(Tris) =>
V=0.4075/(121.14*0.1154) =0.029 L

Answer 2

Simple concept

Equate Number of moles of HCL and TRIS.
(When u neutralize equal moles of acid and base are used
if both have same 'n' factor.)

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