Question

A solution is prepared by mixing 90.0 mL 0.439 M NaCl, 100.0 mL 0.0516 M MgCl2, and 235.0 mL of water. What are the molarities of [Na+], [Mg2+], and [Cl -] in the resulting solution? Please tell me how I would go about solving this.

Answer 1

1) write an equation for the possible reaction
2) determine based on solubility rules if a reaction will occur. meaning if a precipitate will form

Answer 2

It doesn't seem like to me that a reaction will occur, so how do I find the molarities of each individual element?

Answer 3

Do you know the formula for molarity? Can you calculate the number of moles of Na in the original solution of 90ml of 0.439M

Answer 4

Yea, moles solute over liters solution, but Na is bonded with Cl, how do I find Na and not NaCl?

Answer 5

Based on solubility rules what will happen to NaCl in water?

Answer 6

It becomes Na + Cl, right? So do I just used the values given to me for NaCl to find Na?

Answer 7

Or do I convert NaCl to Na?

Answer 8

Yes

Answer 9

For every mole of NaCl there will be one mole of Na

Answer 10

Oh! Since it's moles, we can just say the values are the same?

Answer 11

Correct

Answer 12

So would I just say that the molarity of Na is the same as the given molarity of NaCl?

Answer 13

The original molarity of Na is the same as the original molarity of NaCl. However, the molarity changes; since M = moles solute / L solution, the volume of the original molarity changes since MgCl2 and water is being added.

Answer 14

First calculate the number of moles Na in the original NaCl solution. Please post your answer.

Answer 15

439 moles per 1 liter? And do I find what it is in 90 mL then?

Answer 16

You meant 0.439 mole per L. Yes find what it is in 90 ml. What do you need to convert ml to?

Answer 17

Liters, so it would be something like (0.439 mol Na) x (90mL/1L)?

Answer 18

I'm not sure what you mean.

There are two methods of doing it:

Method 1)

\[\frac{ ml }{ 1 }*\frac{ L }{ ml }*\frac{ mol }{ L }\]

Answer 19

Method 2:

convert ml to L

\[M = \frac{ mole solute }{ L solution }\]

Plug in the value for M and L and algebraically solve for moles

Answer 20

Ok, my teacher likes dimensional analysis/factor labelling/whatever you want to call it, so I would use method 1. I got 0.03951 moles as my answer.

Answer 21

Although that should be 0.040 moles with sig figs.

Answer 22

Correct. So now calculate the new molarity, using 0.03951 moles / (original solution + MgCl2 solution + water)

Answer 23

Keep the 0.03951 in your calculator round to sig figs in your last step/

Answer 24

I would love to, but my teacher tells us to do sig figs for every step, so I'll do it her way.

Answer 25

That's fine with me

Answer 26

So 0.040 moles / 425mL of everything added together is 0.040 moles / 0.425 L of total volume? So 0.094 M?

Answer 27

Your calculation is correct; you are missing a sig fig with 0.094 M.

Answer 28

Oh yea, true. So it's 0.0941.

Answer 29

Correct. You can do the same now for Mg2+ and Cl-. Just be sure to use the mole Cl from NaCl (which is a 1:1 ratio) and mol Cl- from MgCl2 which is a 1:2 mole ratio of 1 mol MgCl2 : 2 mol Cl-. Good Luck

Answer 30

is this high school chemistry or college chemistry?

Answer 31

High school, but accelerated.

Answer 32

One thing: For Cl do I add the moles and liters of it from NaCl and MgCl2 and do a total?

Answer 33

You'll have a much easier in time in college compared to your classmates that did not have accelerated chemistry.

Answer 34

I'm glad to hear it, this is confusing so far.

Answer 35

\[\frac{ (molCl^- fromNaCl)+(molCl^- from MgCl2) }{ L soltn NaCl2 + L soltn MgCl2 + L water }\]

Answer 36

Ok, that makes sense. Thanks for helping, I'll get around to this soon.

Answer 37

The more you practice the less confusing it gets. It starts to make sense a semester to a year later.

Answer 38

I hope so. Ok, going to go eat and then try this after.