Question

What is the OH - in the final solution prepared by mixing 20.0 ml of 0.050 M HCl with 30.0 ml of 0.10 M Ba(OH)2

Answer 1

H+ ions present in HCl solution:  20.0/1000 * 0.050 = 0.001 moles

There are two OH- ions in a molecule of Ba (OH)2.
OH- ions released in the Barium Hydroxide solution =
        = 2 * 30.0/1000 * 0.10 = 0.006 moles

So 0.001 moles of H+ and OH- combine to form water.

So 0.005 moles of OH- remain. So the molarity of [ OH- ]
            = 0.005 moles / (20+30) ml   =0.100 M

Answer 2

Answer:

H+ ions present in HCl solution:  20.0/1000 * 0.050 = 0.001 moles

There are two OH- ions in a molecule of Ba (OH)2.

OH- ions released in the Barium Hydroxide solution =

       = 2 * 30.0/1000 * 0.10 = 0.006 moles

So 0.001 moles of H+ and OH- combine to form water.

So 0.005 moles of OH- remain. So the molarity of [ OH- ]

           = 0.005 moles / (20+30) ml   =0.100 M