Question

How many milliliters of a .239M HCl solution are needed to neutralize 309mL of a .0407 M Ba(OH)2 solution?

Answer 1

@CShrix

Answer 2

@Compassionate

Answer 3

@sweetburger

Answer 4

@dan815

Answer 5

@radar

Answer 6

@pooja195

Answer 7

@zepdrix

Answer 8

Can you write the balance equation between HCl & Ba(OH)2?

I think you meant to write milliliters.

Answer 9

I did. Sorry.

Answer 10

I fixed it.

Answer 11

@LeibyStrauss Ba(OH)2 + 2HCl >> BaCl2 + 2H2O

Answer 12

Good. How many moles Ba(OH)2 are there?

Answer 13

1 mol

Answer 14

In 309 mL how many moles are there?

Answer 15

.0407 mols

Answer 16

0.0407 is M which is molarity.

Do you know what molarity represents?

Answer 17

mols/ liters

Answer 18

Good. So how many mole of Ba(OH)2 are in 309 mL?

Answer 19

.0407mols/.309L

Answer 20

Correct, so how many moles of HCl are needed?

Answer 21

.239 mols/x L

Answer 22

I made a mistake earlier. moles is not 0.0407 / 0.309. Let's go back to Ba(OH)2, how many moles of Ba(OH)2 are in .309 L?

Answer 23

1 mol?

Answer 24

What is the number of moles in 0.309L of 0.0407M of Ba(OH)2?

Answer 25

.0125763 mol

Answer 26

Correct. So, how many moles of HCl are needed to react with 0.125763 mol Ba(OH)2?

Answer 27

Would I figure that out the same way?

Answer 28

You need to go back to the balanced equation you wrote and see how many moles of HCL are needed for every mol of Ba(OH)2

Answer 29

2 mol

Answer 30

Correct. So you need 2 times as much moles of HCl than Ba(OH)2. How many moles of HCL are needed?

Answer 31

.478 mol

Answer 32

What calculation did you do?

Answer 33

2*.239 since you need 2 times as much.

Answer 34

Where's the 0.239 from? You had 0.125763 mol Ba(OH)2?

Answer 35

I was using the .239 M from the HCl.
So it would be .251526 mol

Answer 36

We don't use the M from HCl yet. If we have 0.125763 mol Ba(OH)2 how many moles HCl is needed?

Answer 37

.251526 mol

Answer 38

Good. How many L of 0.239 M HCl is 0.251526 mol?

Answer 39

.60114714 L

Answer 40

You originally wrote .0125763 mol Ba(OH)2 which is correct. I mistakenly typed 0.125763. Using the correct number of 0.0125763 you need to to re-calculate moles HCl

Answer 41

.025146 mol

Answer 42

Ok. So how many L of 0.239 M HCL will have .025146 mol HCl?

Answer 43

Would I need to divide or multiple for this?

Answer 44

\[M = \frac{ mol }{ L } => 0.239 = \frac{ 0.0251526 }{ L }\]multiply both side s by L

0.239 L = 0.0251526

Solve for L

Answer 45

.1052410042 L

Answer 46

Now you need to convert it to mL and you're done

Answer 47

I got 105.24

Answer 48

Would I leave it at 105 mL for sig figs?

Answer 49

Yes

Answer 50

Thanks!

Answer 51

Could you check the answer to another problem, please? If you have time.

Answer 52

If it's similar to this. Can you post it as a new post?

Answer 53

I think so. Will do.