Astudent titrates 0.139 g of an unknown monoprotic weak acid to the equivalence point with 44.6 ml of 0.100 m naoh (aq). what is the molar mass of the weak acid?

The  molar  mass  of  the  weak   acid  is  calculated  as  follows

by use of  an  example  of  monoprotic  weak  such  as CH3COOH   reacting  with  NaOH

that  is NaOH + CH3COOH  = CH3COONa + H2O

calculate  the moles  of NaOH  used
=  molarity  x  volume/1000

= 44.6  ml  x  0.100/ 1000  = 4.46  x10^-3  moles

by  use of mole  ratio  between NaOH  to monoprotic  acid which  is1:1  the moles  of   monoprotic  acid  is  also  4.46  x10^-3 moles

molar  mass  is therefore = mass/  moles = 0.139 g/ (4.46 x10^-3) =  31.17 g/mol

The answer is d because i just took the test