Question

If 18.43 ml of a 0.1274 m naoh solution is needed to reach the first equivalence point in titration of a 0.2700g sample of amino aci, what is the molecular weight of the acid

Answer 1

Moles of NaOH required to reach the first equivalence point

= $18.43mL*\frac{1L}{1000mL} *\frac{0.1274molNaOH}{1L}  =0.00235molNaOH$

0.00235 mol NaOH is required to reach the first equivalence point when NaOH is titrated with an amino acid.

So the moles of NaOH = Moles of amino acid in the solution

Moles of amino acid = 0.00235 mol amino acid

The given mass of amino acid =0.2700g

Calculating the molar mass of amino acid from mass and number of moles:

$\frac{0.2700g}{0.00235mol} =114.9g/mol$