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.

Question

anonymous · Answer

$$Ba(OH )_{2}$$

anonymous · Answer

is it asking to only find the molarity of the OH or the entire Ba(OH)2?

anonymous · Answer

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

anonymous · Answer

did u get the molarity of 0.1167?

anonymous · Answer

I didn't do it, whats the molar mass?

anonymous · Answer

171.33

anonymous · Answer

yea thats what i got also

anonymous · Answer

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 · Answer

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 · Answer

$$.1167 moles Ba(OH)_2 (\frac{2 mole OH}{1 mole Ba(OH)_2})=$$

anonymous · Answer

ok

anonymous · Answer

Sorry I missed that when i first told you how to solve it.

anonymous · Answer

ok cool, do you mind me giving you the next part of the question?

anonymous · Answer

whats the next part?

anonymous · Answer

a student took 20ml of the previous solution and added 50ml of water. Calculate the new hydroxide ion concentration

anonymous · Answer

now i know you use m1v1=m2v2

anonymous · Answer

right?

anonymous · Answer

Yea use that

anonymous · Answer

so (0.02L)(0.233)=(0.05L)(x) then solve for x?

anonymous · Answer

Yes that looks like it should work to me.

anonymous · Answer

ok last question

anonymous · Answer

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 · Answer

so [H]x0.0932=1.00x10^(-14)

anonymous · Answer

[H]=1.073x10^(-13) right?

anonymous · Answer

so then the pH is -log(1.073x10^(-13))

anonymous · Answer

12.97?

anonymous · Answer

you got me on this on, i'm stumped.

anonymous · Answer

ok thanks anyway?

anonymous · Answer

sorry i did not mean to put the question mark lol

anonymous · Answer

thank you

anonymous · Answer

No problem sorry i couldn't help with that last part.