[stev3n]

How many mL of 0.25M HCl(aq) are needed to neutralize a solution of 99.50mL of 0.15M barium hydroxide?
2HCl + Ba(OH)2 -> 2H2O + BaCl2

.0995L * .15M = .014925mol Ba(OH)2

.014925 mol Ba(OH)2 (2 mol HCL/1 mol Ba(OH)2) = .02985 mol HCl

.02985 mol HCl / .25 M = 0.1194L or 119.4mL

and now for the part I can't figure out...

Calculate the molarity of the chloride ion ion present in the solution that results from mixing the proper volume of HCl (aq) with the solution of barium hydroxide specified in the preceding problem.

I figure that I have to use stoichimoetry to get this, but I'm just unsure of how to start.

My best guess is something along the lines of (working backwards)..

(2 mol Cl/1 mol BrCl2) (1 mol BrCl2 / 2 mol HCl)

but I really have no idea. Any help would be appreciated. Thank you

[Borek]

First point OK.

Not BrCl₂ but BaCl₂. But it doesn't matter (much).

How many moles of HCl have you used?

What is the final volume of the solution?

[stev3n]

BrCl2 was just a typo.

Okay so I take it this is a limiting reagent equation. So since I only need so many ml of HCL that means I used the original .02985mol HCl.

As far as the volume of the final solution, since no matter can be lost, it would be 119.4mL HCl + 99.50mL BaCl2 = 218.9mL or .2189L.

M=mol/L = .02985mol/.2189L = .13636.

Thanks man.

[Borek]

Okay so I take it this is a limiting reagent equation.

It is not. You have calculated exactly stoichiometric amount of hydrochloric acid needed.

So since I only need so many ml of HCL that means I used the original .02985mol HCl.

Exactly.

As far as the volume of the final solution, since no matter can be lost, it would be 119.4mL HCl + 99.50mL BaCl2 = 218.9mL or .2189L.

Typo again - should be 99.50 mL of Ba(OH)₂

M=mol/L = .02985mol/.2189L = .13636

That's OK. Just don't usee too many significant digits.