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