How many milliliters of 0.125 M HCl are needed to completely neutralize 43.0 mL of 0.100 M Ba(OH)2 solution? Express the volume in milliliters to three significant digits.

Question

It is solution stoich. I have tried 5 times and I still cant seem to get the formula right, some help would really be appreciated.

Answer 1

Hello, Ava,

We need to start with a balanced equation for the neutralization.

2HCl + Ba(OH)₂ = BaCl₂ + 2H₂O

Each molecule/mole of Ba(OH)₂ requires 2 molecules(or moles) of HCl.

Now let's find the number of moles we have of Ba(OH)₂. M, or Molar, is defined as moles/liter. 1 liter of 0.100 M Ba(OH)₂ contains 0.100 moles Ba(OH)₂. But we only have 43 ml. To determine the number of moles that represents, multiply the concentration 0.125 M (or moles/liter) times the volume, in liters. 43.0 ml is 0.043 liters, so (0.100 moles/liter)*(0.043 liters) = 0.0043 moles of Ba(OH)₂.

We already decided that we need 2 moles of HCL for every mole of Ba(OH)₂. So to neutralize the Ba(OH)₂ we need twice that many moles of HCl. ((2 moles HCl)/(1 mole Ba(OH)₂))*(0.0043 moles Ba(OH)₂ = 0.0086 moles of HCl.

The HCl comes as a 0.125M solution. We have 0.125 moles HCl per each liter. To find the volume that contains the 0.0086 mole needed, Use (0.125 moles/liter) * (X liter) = 0.086 moles, where X will be the volume needed for 0.0086 moles of HCl. I get 0.0688 liters, or 68.8 ml of HCl. (3 sig figs).

I hope this will help,

Bob