# What volume required?

Integral0
What volume of .273 Molarity HCl is required to react completely with 50.00 mL of .167 Molarity of Barium Hydroxide?

2 HCl + Ba(OH)2 -> 2 H20 + BaCl

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I am confused about this problem because my profe told me that you need to double the amount of Hydrogen b/c there are 2 moles of HCl. Can someone explain this?

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Also, my profe solved the problem by this method:

50.00 mL x L/1000 x .167 mol = 8.35 x 10^-3 mol Ba(OH)2 which means Ba is 8.35 x 10^-3 mol. OH is 8.35 x 10^-3 mol.

This is where I get confused.

Then he took 2 mol HCl x (8.35 x 10^-3 OH) x (1L/.273) mol x 10^3 mL of HCl.

I don't understand why he used HCl with OH and how he canceled the OH out to get the HCl.

Thanks for the help!

## Answers and Replies

Staff Emeritus
Gold Member
Originally posted by Integral0
What volume of .273 Molarity HCl is required to react completely with 50.00 mL of .167 Molarity of Barium Hydroxide?

2 HCl + Ba(OH)2 -> 2 H20 + BaCl

--------------------------------------------

I am confused about this problem because my profe told me that you need to double the amount of Hydrogen b/c there are 2 moles of HCl. Can someone explain this?
Hi there Integral0!

For your first question, I'll explain it step by step.

First, calculate how many mmols of Ba(OH)2 you have got.
50 ml * 0.167 mmol/ml = 8.35 mmol

Now, the equation says the following: for every molecule of Ba(OH)2 you have got, you'll need TWO molecules of HCl to react with it.

That is why you have to multiply the amount of Ba(OH)2 (don't devide it by two, otherwise the formula would've read 0.5 HCl + Ba(OH)2 ! )

So: 8.35 * 2 = 16.7 mmol of HCl is needed
Your concentration is 0.273 mmol/ml
16.7 / 0.273 = 61.2 ml HCl

Does that make sense to you?

Staff Emeritus
Gold Member
Originally posted by Integral0
Also, my profe solved the problem by this method:

50.00 mL x L/1000 x .167 mol = 8.35 x 10^-3 mol Ba(OH)2 which means Ba is 8.35 x 10^-3 mol. OH is 8.35 x 10^-3 mol.

This is where I get confused.

Then he took 2 mol HCl x (8.35 x 10^-3 OH) x (1L/.273) mol x 10^3 mL of HCl.

I don't understand why he used HCl with OH and how he canceled the OH out to get the HCl.

Thanks for the help!
Wow :) your proffessor knows how to write a complex equation :)

I always like to write things step by step: what do I have got and what do I need?

I also like to keep things simple by not calculating with the 10^-3 as you saw, I didn't use it at all to solve the problem, but came to the same answer.

A molarity is expressed as mole per liter. But you can change it into anything you like, millimole per milliliter, micromole per microliter. As long as the units stay the same at both sides!!

Since the units are displayed in milliliters, I decided to calculate the molarities in millimoles per milliliters (mmol/ml).

As for the confusion why he used HCl with OH, that must have been a typo. He took HCl with OH2! That makes sense, since those are reacting with eachother.

And as the formula dictates, for every molecule of OH2 that you have got, you'll need TWO molecules of HCl.

Integral0
thank you

thank you Monique

Staff Emeritus