. Boric acid can be used to neutralize bases. The equation for the reaction is: H3BO3(s) + 2OH - (aq) → 2H2O.

What volume of 0.216 M Ba(OH)2 can be neutralized by 10.00 g of H3BO3?

HOW do I solve it?? I have a test tom!
thanks!

Do you have the correct eqn? H3BO3 is unique as a protic acid. I believe the neutralization eqn is
B(OH)3 + OH^- → [B(OH)4]^- that is 1:1
61.8338 g require 1 mol of OH^-
10 g .: require (1.00/61.83)×10 = 0.1617 mol of OH^-
0.216 M Ba(OH)2 is 0.432 M in OH^-
0.432×V = 0.1617
= 0.374 L
If you think H3BO3 requires 2 mol of OH^- then change the 1 to 2 and answer
will be 0.749 L