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anonymous
 4 years ago
A solution is made by dissolving 5g of Ba(OH)2 to a total volume of 250ml. Calculate the molarity of hydroxide ion in this solution.
anonymous
 4 years ago
A solution is made by dissolving 5g of Ba(OH)2 to a total volume of 250ml. Calculate the molarity of hydroxide ion in this solution.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is it asking to only find the molarity of the OH or the entire Ba(OH)2?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I would do this in 2 steps first calculate how many moles are in 5 grams of baoh2 and then figure out how many moles it is per per ml then multiply that times 1000

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0did u get the molarity of 0.1167?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I didn't do it, whats the molar mass?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea thats what i got also

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so isnt that the molarity of the entire Ba(OH)2? why is it saying to find the molarity of the hydroxide ion? are they the same question?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok take molarity of baoh2 and do a little stoichometry using the conversion factor 1 mole Ba(OH)2 = 2 mole OH Can you pull that one off?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[.1167 moles Ba(OH)_2 (\frac{2 mole OH}{1 mole Ba(OH)_2})=\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry I missed that when i first told you how to solve it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok cool, do you mind me giving you the next part of the question?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0a student took 20ml of the previous solution and added 50ml of water. Calculate the new hydroxide ion concentration

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now i know you use m1v1=m2v2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so (0.02L)(0.233)=(0.05L)(x) then solve for x?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes that looks like it should work to me.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i have two equations: pH=log[H] and [H][OH]=1.00x10^14. (The symbol "[ ]" indicates concentration/molarity). Using the previous the last molarity of OH (0.0932) to calculate the pH

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so [H]x0.0932=1.00x10^(14)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0[H]=1.073x10^(13) right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so then the pH is log(1.073x10^(13))

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you got me on this on, i'm stumped.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry i did not mean to put the question mark lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No problem sorry i couldn't help with that last part.
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