I am confused about where I am going wrong here...

Question

Suppose that 1.331 g of impure barium hydroxide is dissolved in enough water to produce 250. mL of solution and that 35.0 mL of this solution is titrated to the stoichiometric point with 17.6 mL of 0.0935 M HCl(aq). What percentage of the impure sample was barium hydroxide?

I found the concentration of the 35 mL Ba(OH)2 soln to be .047M and I know that there is a 1:2 ratio for Ba(OH)2 and HCl. I attempted to find the amount of moles of HCl (which I thought was .0016) and dividing that by 2 to get .000823 mol Ba(OH)2. I then multiplied that by MW 171.34 g/mol to get .141 g. I then divided this by the total given (1.331g) and found my percentage to be 10.59, which is incorrect.  I've tried doing this a few times, playing around with the values and methods but can't get it right. As always, help is much appreciated!

DrBob222 · Accepted Answer

I think the error is in the first step. Since the sample being analyzed is not pure Ba(OH)2, the molarity of an impure sample doesn't mean anything. Here is the way I would approach it; all the steps you already know.
1. I like to work in millimiles so let me do that and if you don't like it just divide mL by 1000 and mmoles by 1000. (mL x M = millimoles; M = mmoles/mL.

mmoles HCl = 17.6 mL x 0.0935 = about 1.6

mmoles Ba(OH)2 = 1.6 x 1/2 = about 0.8 because the equation is
2HCl + Ba(OH)2 ==> etc

grams Ba(OH)2 = moles x molar mass = about 0.0008 x 171.34 = about 0.14g

percent = (mass Ba(OH)2/mass sample)*100

mass sample = 1.331 x (35 mL/250 mL) = about 0.19 grams.

%Ba(OH)2 = (about 0.14 g Ba(OH)2/about 0.19)*100 = about 75% Ba(OH)2.
Check my thinking. Check my arithmetic.

[Note: just for your information, I do all of this in one step.
(mL x N x milliequivalent weight/mass sample *100 = % but the IUPAC, in all of it's wisdom, has declared equivalent weight obsolete and it isn't taught anymore. Gravimetric factors have been declared obsolete, also, and aren't taught anymore. I see some mention (a sentence or two) in some texts about normality but nothing about gravimetric factors.