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 have tried to solve this problem twice but haven't been successful. I don't need the final answer; I just want to know how to set it up and where to start, as it seems I'm not approaching it correctly. The dissociation equation given is Ba(OH)₂ → Ba²⁺ + 2OH⁻. Thank you for your help, and please show your work.

Anonymous · 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 litres
so number of moles of solute = molarity X volume of solution in litres
so number of moles of Ba(OH)2 = 0.00075 X 1 = 0.00075 moles
now 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 M
also 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 M
edit- thanks for correcting me .....you are the best user in this section

Prof. Mozelle Willms · 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

Anonymous

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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