BaOH2 will dissociate into Ba 2 and OH

Ba(OH)2 will dissociate into Ba 2+ and OH- ions when placed in solution. The availability of these mobile ions in the solution will cause electrical conductivity to be ‘high’ As H 2 SO 4 is added to the solution it will neutralize the Ba(OH)2 producing Ba. SO 4 which precipitates out as it is insoluble in water. As a result there will be fewer mobile ions left to conduct electricity. So the conductivity will decrease.

The reaction between H 2 SO 4 and Ba(OH)2 is given as: H 2 SO 4 + Ba(OH)2 → Ba. SO 4 + 2 H 2 O From this we find that one mole of H 2 SO 4 will neutralize one mole of Ba(OH)2 So adding equal number of moles of H 2 SO 4 will completely neutralize the original Ba(OH)2. The resulting Ba. SO 4 is not soluble in water, so it will precipitate out. The end result will be lower conductivity as there will be ‘almost’ no mobile ions left. (cause of Ksp of course ) The presence of excess H 2 SO 4 provides mobile ions in the solution because it dissociates into H+ and HSO 4 - ions. As a result the solution conductivity continually increases.

Yes, because Ba. SO 4 is insoluble in solution; at the equivalence point the conductivity will be low. (but not zero. . it’s the Ksp thing ) HCl + Na. OH → Na. Cl + H 2 O 27. 55 millilitres of 0. 0213 molar HCl 0. 0213 mol HCl L x 0. 02755 L = 5. 868 x 10 -4 mol HCl Since this is the equivalence point. . . 5. 868 x 10 -4 mol Na. OH 0. 050 L = 0. 0117 M Na. OH

2) 27. 55 m. L of 0. 0213 M HCl

1) Conductivity (μS) Equivalence point volume of H 2 SO 4 (m. L)