Question

Calculate the molarity of each type of ion remaining in solution after 30.0 ml of 4.00 M HBr is mixed with 100.0 ml of 1.00 M Ca(OH)2 and 50.0 ml of water. 2HBr + Ca(OH)2 --> CaBr2 + 2H2O

Can someone explain how to solve this problem?

Answer 1

This is a long question I think. First you'll have to find which one is the limiting reagent and find the excess of the other substance. Can you do that?

Answer 2

My professor already gave us the steps and the answer but I didn't understand why and how he solved it.

Answer 3

Can you post the steps as well then and tell me which part you don't get?

Answer 4

Here's the pic of the steps, hopefully its readable....I don't understand why he choose to subtract 50 and why did he only add 50 to KI and not H2O. He said this was the shortcut but this shortcut is kinda confusing, unless there's another way of solving it.

Answer 5

Ignore the sentence that says "In nature Chlorine is found..." It was part of another problem.

Answer 6

do you know how to find moles?

you should start by doing that for all the species added to the reaction mix

Answer 7

Wait. That looks like a different reaction O.o

Answer 8

But either way:
1) calculate number of moles of anything you have
2) see which one is the limiting reagent
3) find the excess number of moles of the other reagent

I think that's how you do it

Answer 9

oops that is another reaction:/ but either way I didnt understand this one which is exactly the one above....

Answer 10

So you don't get the one you originally posted, and your professor did not show you how to do it?

Answer 11

https://smartsite.ucdavis.edu/access/content/group/b5a9176e-40d9-48f5-a473-7c9b26ee9e27/EXAMS/EXAM%201/PRACTICE%20EXAMS/KEY%20EXAM%201%20F2008.pdf

Answer 12

He kinda showed us how to solve it but me and my roommate didn't understand how and why he did the stuff...

Answer 13

Is there another way of solving this problem though? Thats much easier to understand?

Answer 14

I can't access that. I need a log in. But maybe someone can help you. Good luck

Answer 15

My professor subtracted 120 from HBr and CaOH2 and added 120 to CaBr2. Let me try to get a screenshot of what he did

Answer 16

Sure. :)

Answer 17

Its question number 17 on the pdf

Answer 18

Ah, got it!

The first steps were right.

What he did next was to subtract 120 mmol from HBr. Therefore, the amount of Ca(OH)2 that will be used to combine with 120 mmol HBr is

120 mmol HBr x [1 Ca(OH)2/2 HBr] = 60 mmol Ca(OH)2.

So subtract 60 mmol to Ca(OH)2 (giving 100 mmol - 60 mmol = 40 mmol).
Particularly, since all 120 mmol of HBr are used up, then the remaining will be 0.

Now, the the amount of product CaBr2 can now be computed, thus

60 mmol Ca(OH)2 x [1 CaBr2/1 (Ca(OH)2] = 60 mmol CaBr2. (So add 60 to CaBr2).

You can disregard H2O since no ions can be seen in this compound.

Summarizing,
Before After
HBr 120 0
Ca(OH)2 100 40
CaBr2 0 60

Answer 19

Now, 1 mmol of Ca(OH)2 has 1 mmol Ca2+ and 2 mmol OH-.
So, 40 mmol of Ca(OH)2 yields 40 mmol Ca2+ and 80 mmol OH-.
Also, 1 mmol of CaBr2 has 1 mmol Ca2+ and 2 mmol Br-.
So, 60 mmol of CaBr2 yields 60 mmol Ca2+ and 120 mmol Br-.

Summarizing,

Ca2+ = 40 + 60 = 100 mmol
OH- = 80 mmol
Br- = 120 mmol = 1.2 x 10^2 mmol.