## 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.

1. anonymous

$Ba(OH )_{2}$

2. anonymous

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

3. 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

4. anonymous

did u get the molarity of 0.1167?

5. anonymous

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

6. anonymous

171.33

7. anonymous

yea thats what i got also

8. 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?

9. 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?

10. anonymous

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

11. anonymous

ok

12. anonymous

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

13. anonymous

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

14. anonymous

whats the next part?

15. anonymous

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

16. anonymous

now i know you use m1v1=m2v2

17. anonymous

right?

18. anonymous

Yea use that

19. anonymous

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

20. anonymous

Yes that looks like it should work to me.

21. anonymous

ok last question

22. 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

23. anonymous

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

24. anonymous

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

25. anonymous

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

26. anonymous

12.97?

27. anonymous

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

28. anonymous

ok thanks anyway?

29. anonymous

sorry i did not mean to put the question mark lol

30. anonymous

thank you

31. anonymous

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