200.0 mL of 3.85 M HCl is added to 100.0 mL of 4.6 M barium hydroxide. The reaction goes to completion. What is the concentration (in mol/L) of excess H+ or OH- that remains in solution?

2.387 mol/L

Explanation:

The reaction that takes place is:

First we calculate how many moles of each reagent were added:

460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.

Now we calculate how many moles of Ba(OH)₂ reacted, by converting the total number of HCl moles to Ba(OH)₂ moles:

This means the remaining Ba(OH)₂ is:

There are two OH⁻ moles per Ba(OH)₂ mol:

Finally we divide the number of OH⁻ moles by the total volume (100 mL + 200 mL):

So the answer is 2.387 mol/L

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