A solution 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