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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?

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Answer:

2.387 mol/L

Explanation:

The reaction that takes place is:

  • 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

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

  • HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl
  • Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂

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:

  • 203.85 mmol HCl * [tex]\frac{1mmolBa(OH)_{2}}{2mmolHCl}[/tex]= 101.925 mmol Ba(OH)₂

This means the remaining Ba(OH)₂ is:

  • 460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂

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

  • OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻

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

  • 716.15 mmol OH⁻ / 300.0 mL = 2.387 M

So the answer is 2.387 mol/L

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