How many milliliters of a 0.197 M HCl solution are needed to neutralize 319 mL of a 0.0335 M Ba(OH)2 solution?

Answer 1

Always begin by writing the correctly balanced equation for the reaction that is taking place:

2HCl + Ba(OH)₂ ===> BaCl₂ + H₂O

From this equation, we see that it will take TWO moles of HCl for each one mole of Ba(OH)₂. So, we will first determine how many moles of Ba(OH)₂ are present. From that, we can easily find the moles HCl needed (it will be twice the number of moles of Ba(OH)₂. Finally, knowing the moles of HCl needed, we can find the volume needed. Here goes...

moles Ba(OH)₂ present = 319 ml x 1 L/1000 ml x 0.0335 mol/L = 0.010687 moles

moles HCl needed = 0.010687 mol Ba(OH)₂ x 2 mol HCl/mol Ba(OH)₂ = 0.02136 moles

To find the volume of 0.197 M HCl needed to provide 0.02136 moles HCl, we proceed as follows:

(x L)(0.197 moles/L) = 0.02136 moles

x = 0.02136/0.197 = 0.1084 liters = 108.4 mls = 108 mls (to 3 significant figures)