Asolution of 0.393 m ba(oh)2 required 32.24 ml of a 0.227 m solution of hn03 calculate the original volume of the ba(oh)2 soluion

The  original  volume    of  the  Ba(OH)2  solution  is calculated  as  follows

find the  moles   of HNO3  used

Ba(OH)2  +  2HNO3  =Ba(NO3)2  +2H2O

calculate  the  moles   of   HNO3  used  =molarity  x  volume
= 32.24  x0.227=  7.32   moles

by  use of  mole  ratio  between  Ba(OH)2 to HNO3  which  is   1:2 the moles  of  Ba(OH)2  is  therefore  =  7.32  x1/2 =3.66   moles

volume of Ba(OH)2   is  therefore = (3.66  /0.393= 9.3 ML  of Ba(OH)2

The answer is probably c. xylem