Calculate the volume of 6.83×10-2 M barium hydroxide required to neutralize 14.8 mL of a 0.274 M hydroiodic acid solution.

Question

Please help with this problem. Can you also provide a step-by-step way of solving it?

Answer 1

Balanced equation:

Ba(OH)₂ + 2HCl ==> H₂O + BaCl₂

First, find MOLES HCl present:

14.8 ml x 1L/1000 ml x 0.274 moles/L = 0.004055 moles HCl

Next, using the stoichiometry of the balanced eq. (1Ba(OH)₂ : 2 HCl), find moles Ba(OH)₂ needed:

0.004055 moles HCl x 1 mol Ba(OH)₂/2 moles HCl = 0.002028 moles Ba(OH)₂

Finally, calculate volume of 6.83x10^-2 M Ba(OH)₂ needed to provide 0.002038 moles:

(x L)(6.83x10-2 mol/L) = 0.002038 moles

x = 0.02969 L = 29.7 mls (to 3 significant figures)