anonymous
  • anonymous
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.
Chemistry
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[Ba(OH )_{2}\]
anonymous
  • anonymous
is it asking to only find the molarity of the OH or the entire Ba(OH)2?
anonymous
  • anonymous
I 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

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anonymous
  • anonymous
did u get the molarity of 0.1167?
anonymous
  • anonymous
I didn't do it, whats the molar mass?
anonymous
  • anonymous
171.33
anonymous
  • anonymous
yea thats what i got also
anonymous
  • anonymous
ok 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
  • anonymous
ok 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
  • anonymous
\[.1167 moles Ba(OH)_2 (\frac{2 mole OH}{1 mole Ba(OH)_2})=\]
anonymous
  • anonymous
ok
anonymous
  • anonymous
Sorry I missed that when i first told you how to solve it.
anonymous
  • anonymous
ok cool, do you mind me giving you the next part of the question?
anonymous
  • anonymous
whats the next part?
anonymous
  • anonymous
a student took 20ml of the previous solution and added 50ml of water. Calculate the new hydroxide ion concentration
anonymous
  • anonymous
now i know you use m1v1=m2v2
anonymous
  • anonymous
right?
anonymous
  • anonymous
Yea use that
anonymous
  • anonymous
so (0.02L)(0.233)=(0.05L)(x) then solve for x?
anonymous
  • anonymous
Yes that looks like it should work to me.
anonymous
  • anonymous
ok last question
anonymous
  • anonymous
i 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
  • anonymous
so [H]x0.0932=1.00x10^(-14)
anonymous
  • anonymous
[H]=1.073x10^(-13) right?
anonymous
  • anonymous
so then the pH is -log(1.073x10^(-13))
anonymous
  • anonymous
12.97?
anonymous
  • anonymous
you got me on this on, i'm stumped.
anonymous
  • anonymous
ok thanks anyway?
anonymous
  • anonymous
sorry i did not mean to put the question mark lol
anonymous
  • anonymous
thank you
anonymous
  • anonymous
No problem sorry i couldn't help with that last part.

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