Question

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

Answer 1

Answer:(3.66 /0.393= 9.3 ML of Ba(OH)2

Explanation: 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

Answer 2

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