chemistry: What are the concentrations of OH– and H in a 0.00075 M solution of Ba(OH)2 at 25 °C? Assume comple?

Question

What are the concentrations of OH– and H in a 0.00075 M solution of Ba(OH)2 at 25 °C? Assume complete dissociation.i tried twice and i can't figure it out...i DONT want the answer.. I just want to knwo how to set it up and where to start...because obviously i'm not doing it right. they just gave me Ba(OH)2 --> Ba + 2OHthank you! please show work...

Johathan Wehner · Accepted Answer

Ba(OH)2 -----> Ba2+ + 2OH-since in the question there is no mention about the volume of solution so i am assuming it as 1 L solution now molarity = number of moles of solute/volume of solution in litresso number of moles of solute = molarity X volume of solution in litresso number of moles of Ba(OH)2 = 0.00075 X 1 = 0.00075 molesnow 1 mole of Ba(OH)2 gives 2 moles of OH-so 0.00075 moles of Ba(OH)2 will give 0.00075 X 2 = 1.5 X 10^-3 moles of OH-molarity of OH- = 1.5 X 10^-3 / 1 = 1.5 X 10^-3 Malso as Kw = [H+] [OH-]so 10^-14 = [H+] X 1.5 X 10^-3[H+] = 10^-14/1.5 X 10^-3 = 6.667 X 10^-12 Medit- thanks for correcting me .....you are the best user in this section

Johathan Wehner · Answer

You have concentration and you want concentration as an answer: The volume of solution does not matter. Ba(OH)2 ↔ Ba2+ + 2OH- 0.00075M Ba(OH)2 = 0.000752 = 1.510^-3M OH- [OH-] [H+] = 10^-14[H+] = 10^-14/[OH-] = [H+] = 10^-14 / (1.5*10^-3) [H+] = 6.67*10^-12M

Johathan Wehner

chemistry: What are the concentrations of OH– and H in a 0.00075 M solution of Ba(OH)2 at 25 °C? Assume comple?

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